 pharmaceutical calculations questions and answers

Note:: First try your best to solve these pharmaceutical calculations. If you find them difficult you can consult the step-by-step solutions at the end of the MCQ questions.

76) What is the volume, in milliliters, of 85.7% w/w lactic acid (with a specific gravity of 1.19) required to prepare 120 mL of a 10% w/v lactic acid solution?
a- 7.6 mL
b- 11.8 mL
c- 14.5 mL
d- 19.4 mL
e- 22.6 mL

Solution:

This is a pharmaceutical calculation problem related to the dilution of acids.

Concentrated acids are expressed as w/w

Diluted acids are expressed as w/v

To express the (85.7% w/w lactic acid) concentrated acid as diluted;

We can use the following equation to achieve our target;

w/v = w/w x density ________ equation 1

We are not given the density of lactic acid but the specific gravity of 1.19 is given and we can calculate density from it, as

specific gravity = density of lactic acid / density of water at 4°C

The density of water at 4°C is 1 g/mL.

density of lactic acid = specific gravity / density of water at 4°C

density of lactic acid = 1.19 / 1 = 1.19 g/mL

Put this density into equation 1, we get;

w/v = 85.7% x 1.19 = 101.983 w/v

Now we can solve this problem by using the formula

Q1C1 = Q2C2

Q1 = the quantity (volume) of the starting solution = unknown
C1 = the concentration of the starting solution = 101.983 w/v
Q2 = the desired quantity (volume) of the diluted solution = 120 mL
C2 = the desired concentration of the diluted solution = 10% w/v

Putting the values into the formula;

x mL (101.983 w/v) = 120 mL (10% w/v)

x = 120 mL (10% w/v) / (101.983 w/v)

x = 11.76 mL

Therefore, approximately 11.8 mL of 85.7% w/w lactic acid is needed to make 120 mL of 10% w/v lactic acid.

Thus, the correct option is b.

77) How many milliosmoles (mOsm) are present in 1 liter of sodium chloride injection? (Molecular weight of sodium chloride = 58.5)
a- 0.308 mOsm
b- 3.08 mOsm
c- 30.8 mOsm
d- 308 mOsm
e- 3080 mOsm

Solution:

An osmole is a unit of measurement that represents the number of particles in a solution, such as ions or molecules, that can contribute to the osmotic pressure of the solution.

A milliosmole (mOsm) is equal to one-thousandth of an osmole. The formula for milliosmole  is;

mOsm = millimoles * number of species

A mole (mol) is a unit of measurement used to express the amount of a substance. The formula for mole is:

moles = Wt. of a substance in g / molecular weight

A millimole (mmol) is one-thousandth of a mole.

millimoles = (Wt. of a substance in g / molecular weight) × 1000

Now,

Molecular weight of sodium chloride = 58.5

We have to find out the number of milliosmoles present in 1 liter of sodium chloride injection. For that we have to know about the millimoles and number of species of sodium chloride.

The number of species is the number of species per molecule of solute (which is 2 for sodium chloride, since it dissociates into one sodium ion and one chloride ion).

Now to calculate the millimoles;

To find the Wt. of a substance in g, that is the concentration of sodium chloride in the solution;

Since we know that 0.9% NaCl contains 0.9 grams of sodium chloride in 100 mL of injection but we are given 1 liter of sodium chloride injection, so

Wt. of a substance in g = (0.9 / 100) x 1000 = 9 grams

Therefore, 9 grams of NaCl are there in 1 liter of injection.

Putting the values into the formula;

millimoles = ( 9 grams / 58.5 g/mol ) x 1000 = 153.846

Now put the values into the mOsm formula;

mOsm = 153.846 x 2

mOsm = 307.69 mOsm / L

Therefore, 1 liter of sodium chloride injection with a concentration of 0.9% w/v contains about 308 mOsm.

Thus, the correct option is d.

78) What is the concentration of a drug in parts per million if the blood level is 8 μg/dL?
a- 0.08 ppm
b- 0.8 ppm
c- 8 ppm
d- 80 ppm
e- 800 ppm

Solution:

1 part per million or PPM is equal to;

1 part (g) per 1,000,000 parts (mL)

i.e. 1 PPM = 1 / 1000,000

Blood level of the drug is 8 μg/dL, so we need to align the units according to the ppm, so lets convert 8 μg/dL into g/mL

1 gram = 1000,000 μg

1 μg = 1 / 1000,000 grams

8 μg = 8 / 1000,000 grams = 0.000008 g

AND

1 Liter = 10 dL = 1000 mL

1 dL = 100 mL

So,

8 μg/dL is equal to 0.000008 g / 100 mL

Now lets calculate PPM;

0.000008 g / 100 mL = x / 1000,000

x = (0.000008 g / 100 mL) x 1000,000

x = Concentration in PPM = 0.08 ppm

So, the concentration of the drug in parts per million is 0.08 ppm.

Thus, the correct option is a.

79) A prescription for the anti-cancer medication Tamoxifen has been given to a patient. The instructions on the prescription indicate: Tabs p.o. i t.i.d. x 2.5 years. Each tablet of Tamoxifen contains 20 mg of the medication. What will be the total amount of Tamoxifen the patient will consume over a period of five weeks?
a- 420 mg
b- 840 mg
c- 1200 mg
d- 2100 mg
e- 2500 mg

Solution:

The prescription "Tabs p.o. i t.i.d. x 2.5 years" means that the patient should take Tamoxifen one tablet by mouth, thrice daily, for a period of two and a half years.

To calculate how many milligrams of Tamoxifen the patient will take in five weeks, we need to determine the total number of tablets the patient will take in that time period.

There are 7 days in a week, so 5 weeks is equal to 5 x 7 = 35 days.

The patient is instructed to take Tamoxifen tablets thrice daily, so they will take 3 tablets per day.

Therefore, the total number of tablets the patient will take in 5 weeks is:
3 tablets/day x 35 days = 105 tablets

Each tablet contains 20mg of tamoxifen, so the total amount of tamoxifen the patient will take in 5 weeks is:

105 tablets x 20mg/tablet = 2,100 mg

Therefore, the patient will take 2,100 mg of tamoxifen in five weeks.

Thus, the correct option is d.

80) What will be the weight of 420 milliliters of light liquid petrolatum with a specific gravity of 0.81, expressed in grams?
a- 94.4 grams
b- 180.6 grams
c- 320 grams
d- 340.2 grams
e- 400 grams

Solution:

To calculate the weight of 420 milliliters of light liquid petrolatum with a specific gravity of 0.81, expressed in grams, we can use the formula:

mass = volume x specific gravity

Where "mass" is the weight in grams, "volume" is the volume in milliliters, and "specific gravity" is the specific gravity of the liquid.

So, for this problem:

mass = 420 ml x 0.81
mass = 340.2 grams

Therefore, the weight of 420 milliliters of light liquid petrolatum with a specific gravity of 0.81, expressed in grams, is 340.2 grams.

Thus, the correct option is d.

81) What volume of an injectable solution containing 25µg of a drug substance per 0.5 mL would be needed to administer 0.25 mg of the drug substance?
a- 3 mL
b- 5 mL
c- 7 mL
d- 10 mL
e- 15 mL

Solution:

We can use proportion to solve this problem.

Since there are 25µg of drug substance in 0.5 mL of solution, we can set up the following proportion:

25µg        0.25 mg
--------  =  -----------
0.5 mL         x mL

To make this proportion work, lets convert mg into µg;

0.25 mg = 250 µg

So,

25µg        250 µg
--------  =  -----------
0.5 mL         x mL

Now solving for x;

x = (0.5 mL x 250 µg) / 25µg

x = 5 mL

Therefore, we need to administer 5 mL of the injectable solution to provide 0.25 mg (or 250 µg) of the drug substance.

Thus, the correct option is b.

82) What is the required quantity in grams of substance X that needs to be added to a 2000 gm solution containing 10% substance X in order to obtain a 25% solution of substance X?
a- 180 grams
b- 250 grams
c- 400 grams
d- 455 grams
e- 500 grams

Solution:

To prepare a 25% solution of substance X, we need to determine how much pure substance X is required to achieve this concentration.

The given solution contains 10% substance X, which means that in 2000 grams of the solution, there are 10% of substance X and 90% of other components (solvent or additives).

To calculate the amount of pure substance X in the existing solution, we multiply the total weight of the solution by the concentration of substance X:

Amount of substance X = 2000 grams × 10% = 200 grams

Now, we need to determine how much additional pure substance X we need to add to the existing solution to reach a total of 25% concentration.

Let's assume the amount of pure substance X to be added is y grams.

The final solution will have a total weight of 2000 grams (existing solution) + y grams (added substance X). The amount of substance X in the final solution will be the sum of the initial amount and the added amount:

Amount of substance X in the final solution = 200 grams (existing) + y grams (added)

To express this as a percentage, we divide the amount of substance X in the final solution by the total weight of the final solution and multiply by 100:

25% = (Amount of substance X in the final solution / Total weight of the final solution) × 100

Substituting the values, we get:

25% = (200 grams + y grams) / (2000 grams + y grams) × 100

Simplifying the equation:

25(2000 + y) = (200 + y)(100)

50000 + 25y = 20000 + 100y

75y = 30000

y = 400 grams

Therefore, 400 grams of pure substance X should be added to the existing 2000 grams of 10% substance X solution to prepare a 25% solution.

Also this problem can be solved by the Alligation method.

Since substance X is there in pure form, we can take its concentration to be 100%, and thus the highest concentration.

The lowest concentration is 10% and the desired concentration is 25%.

So, the resulting Alligation diagram will be;

100%                 15 parts
25%
10%                   75 parts

It is clear from the Alligation diagram, to produce the desired 25% solution of substance X, we will have to take 15 parts of the pure substance X and 75 parts of the 10% substance X solution and mix them together.

To carry out the problem successfully, we will need to convert the parts into grams.

We know the weight of the 75 parts of the 10% solution which is 2000 grams. We just need to find out the weight of the 15 parts of the pure substance X. So,

75 parts         2000 grams
----------   =   ---------------
15 parts            y grams

Solving for y;

y = ( 15 x 2000 ) / 75

y = 400 grams

Thus, the correct option is c.

83) What is the time duration required for the body to receive a full dose of a drug with a half-life of 72 hours?
a- 3 days
b- 6 days
c- 1.5 weeks
d- 2 weeks
e- 3 weeks

Solution:

The half-life (T1/2) of a drug is the time it takes for the concentration of the drug in the body to decrease by half. In the case of a drug with a T1/2 of 72 hours, it means that after 72 hours, the concentration of the drug in the body will be reduced by half.

To determine how long it takes for the body to receive a complete dose of the drug, we can use the concept of multiple half-lives. Generally, it takes around 5 half-lives for a drug to reach a steady-state concentration or for the body to receive a complete dose.

In this case, since the half-life is 72 hours, we multiply the half-life by 5 to estimate the time required for a complete dose. 72 hours multiplied by 5 gives us 360 hours.

Therefore, it takes approximately 360 hours for the body to receive a complete dose of the drug with a T1/2 of 72 hours. This assumes that the drug is being continuously administered or taken at regular intervals.

Since the options are given in days and weeks, we need to convert the hours into the said units;

Converting 360 hours into days;

360 hours  / 24 hours  = 15 days

And in weeks form, we get

15 days / 7 days = 2.142 weeks

Thus, the correct option is d.

84) Calculate the half-life for a drug which has a rate constant of 0.035 minutes?
a- 0.16 hours
b- 0.22 hours
c- 0.33 hours
d- 0.462 hours
e- 0.56 hours

Solution:

To calculate the half-life of a drug with a rate constant of 0.035 minutes, we can use the formula:

Half-life (t½) = (0.693 / rate constant)

Plugging in the given rate constant:

t½ = 0.693 / 0.035

Solving this equation:

t½ = 19.8 minutes

Since options given are in hours format, we need to convert the half-life of the drug from minutes to hours. For that, divide the half-life value by 60 (since there are 60 minutes in an hour):

19.8 minutes / 60 = 0.33 hours

Therefore, the half-life of the drug is approximately 0.33 hours.

Thus, the correct option is c.

85) What is the concentration of the drug in mcg/ml after diluting 5cc of Lidocaine with a concentration of 0.1mg/cc with water up to the 70cc mark?
a- 5 mcg/mL
b- 7 mcg/mL
c- 10 mcg/mL
d- 14 mcg/mL
e- 15 mcg/mL

Solution:

To calculate the strength of the drug in mcg/ml after dilution, we need to consider the dilution formula;

c1v1 = c2v2

where,

c1 is the initial given concentration,
v1 is the initial given volume,
c2 is the desired concentration, and
v2 is the final volume.

In this case, we know:

c1 = 0.1mg/cc
v1 = 5 cc
c2 = ?
v2 = 70 cc

Now we can plug in the values and solve for c2;

0.1mg/cc x 5 cc = c2 x 70 cc

c2 = 0.007 mg/cc

Now we should convert mg/cc into mcg/ml;

Note: 1cc = 1 mL

So we can say we are to convert mg/mL into mcg/mL, or we just need to convert mg into mcg.

So,

0.007 mg/cc = 7 mcg/mL

Therefore, the strength of the drug in the diluted solution would be 7 mcg/ml.

Thus, the correct option is b.

86) What is the pH of a solution when the ratio of ionized to unionized species of drugs is 1 x 10⁻⁶, given that the pKa is 7?
a- 1
b- 2
c- 3
d- 6
e- 7

Solution:

To determine the pH of a solution with a ratio of ionized to unionized species of drugs of 1 x 10⁻⁶, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-] / [HA])

Given that pKa is 7, we can substitute the ratio of ionized (A-) to unionized (HA) species into the equation:

pH = 7 + log(1 x 10⁻⁶)

Using logarithmic properties, we can simplify this equation:

pH = 7 - 6

pH = 1

Therefore, the pH of the solution would be 1.

Thus, the correct option is a.

87) Using Young's rule, what would be the dosage for a 2-year-old child if the adult dose of a drug is 500mg?
a- 25 mg
b- 46.5 mg
c- 71.43 mg
d- 100 mg
e- 250 mg

Solution:

According to Young's rule, the dosage for a 2-year-old child can be calculated using the formula:

Child's dose = (Age of child in years / (Age of child in years + 12)) * Adult dose

Substituting the values:

Child's dose = (2 / (2 + 12)) * 500 mg

Child's dose = (2 / 14) * 500 mg

Child's dose = 71.428 mg

Therefore, the dosage for a 2-year-old child would be approximately 71.428 mg.

Thus, the correct option is c.

88) How many calories does a 450 cc volume of D5W solution provide?
a- 45 kcal
b- 450 calories
c- 65.8 kcal
d- 76.5 kcal
e- 85 kcal.

Solution:

To calculate the calories provided by 450 cc (or ml) of D5W solution, we need to determine the amount of dextrose in it and then convert it to calories.

D5W (Dextrose 5% in Water) is an intravenous (IV) solution. It contains 5 grams of dextrose (a type of sugar) per 100 milliliters (ml) of water.

1 gram of dextrose provides approximately 3.4 calories.

Given that D5W contains 5 grams of dextrose per 100 ml:

Amount of dextrose in 450 cc of D5W = (5 grams / 100 ml) * 450 ml
= 22.5 grams

Calories provided by 450 cc of D5W = 22.5 grams * 3.4 calories/gram = 76.5 calories

Therefore, 450 cc of D5W solution provides approximately 76.5 calories.

A kilocalorie is another word for what's commonly called a calorie so we can also say 76.5 kcal.

Thus, the correct option is d.

89) How much sodium chloride in milligrams is needed to prepare a 1000 cc volume of a 0.45% normal saline solution?
a- 4.5 mg
b- 45 mg
c- 450 mg
d- 1045 mg
e- 4500 mg

Solution:

To calculate the milligrams of sodium chloride required to prepare 1000 cc (cubic centimeters) of 0.45% normal saline solution, we need to know the definition of "0.45% normal saline."

Normal saline refers to a 0.9% saline solution, which means it contains 0.9 grams of sodium chloride per 100 ml of water.

0.45% normal saline would be half the concentration of a 0.9% saline solution.

0.45% normal saline means it contains 0.45 grams of sodium chloride per 100 ml of water.

Thus, to calculate the milligrams of sodium chloride required to prepare 1000 cc of 0.45% normal saline solution;

Grams of sodium chloride = (0.45 / 100) * 1000 ml

Grams of sodium chloride = 4.50 grams

milligrams of sodium chloride = 4500 mg

Therefore, 4500 milligrams of sodium chloride are required to prepare 1000 cc of 0.45% normal saline solution.

Thus, the correct option is e.

90) Given that each inhalation-dose contains 90 micrograms of albuterol, the albuterol inhaler with 18 mg of albuterol can deliver a total of how many inhalation-doses?
a- 90
b- 180
c- 200
d- 800
e- 1800

Solution:

To determine the number of inhalation-doses that can be delivered from an albuterol inhaler, we need to calculate the total number of inhalation-doses using the given information.

1 mg = 1000 µg

The inhaler contains 18 mg of albuterol, which is equal to;

18 mg * 1000 µg/mg = 18,000 µg

Each inhalation-dose contains 90 µg.

Therefore, the number of inhalation-doses that can be delivered is calculated by dividing the total amount of albuterol in the inhaler by the amount of albuterol in each inhalation-dose:

Number of inhalation-doses = Total albuterol / Albuterol per inhalation-dose

Number of inhalation-doses = 18,000 µg / 90 µg = 200

Therefore, the albuterol inhaler can deliver 200 inhalation-doses.

Thus, the correct option is c.

91) According to federal regulations, transportation workers are not allowed to engage in safety-sensitive roles if their breath alcohol concentration (BAC) exceeds 0.04%. How would this concentration be expressed in milligrams per deciliter (mg/dL)?
a- 4
b- 40
c- 400
d- 20
e- 10

Solution:

This problem is about converting breath alcohol percentage to mg/dL.

The given breath alcohol concentration (BAC) is 0.04%.

0.04% breath alcohol means 0.04 grams of alcohol per 100 mL of exhaled breath.

To convert % to mg/dL, we first convert % to grams / L.

(0.04 grams / 100 mL) * 1000 mL = 0.4 grams / L

Now lets convert L into dL and grams into mg;

1 Liter = 10 deciliters
0.4 grams = 400 mg

So, 0.4 grams / L = 400 mg per 10 dL

Thus 400 mg are there in 10 dL, we now need to find out the concentration per dL.

(400 mg / 10 dL) x 1dL = 40 mg /dL

Therefore, 0.04% breath alcohol concentration corresponds to 40 milligrams of alcohol per deciliter of breath (mg/dL).

Thus, the correct option is b.

92) Express the concentration of an ophthalmic solution as a ratio strength that is preserved with 0.0008% (w/v) of an antibacterial?
a- 1:25,000 w/v
b- 1:50,000 w/v
c- 1:75,000 w/v
d- 1:125,000 w/v
e- 1:250,000 w/v

Solution:

To express the concentration of an ophthalmic solution preserved with 0.0008% (w/v) of an antibacterial as a ratio strength, we need to compare the amount of the antibacterial ingredient to the total volume of the solution.

0.0008% (w/v) means that there are 0.0008 grams of the antibacterial ingredient per 100 milliliters (mL) of the solution.

Now, to express this as a ratio strength, we write the amount of the antibacterial ingredient as the numerator and the total volume of the solution as the denominator:

0.0008 grams : 100 mL

However, it's recommended to simplify the ratio by dividing both the numerator and denominator by the same factor to obtain the simplest ratio.

In this case, we can divide both values by 0.0008 to simplify the ratio:

0.0008 grams ÷ 0.0008 = 1 gram
100 mL ÷ 0.0008 = 125,000 mL

Therefore, the concentration of the ophthalmic solution can be expressed as:

1 gram : 125,000 mL

This is the ratio strength of the ophthalmic solution preserved with 0.0008% (w/v) of the antibacterial ingredient.

Thus, the correct option is d.

93) What is the quantity of dextrose in grams needed to make a 10 L solution of dextrose at a concentration of 1 in 500?
a- 5
b- 10
c- 20
d- 50
e- 100

Solution:

To calculate the amount of dextrose required to prepare a 1 in 500 dextrose solution, we need to determine the concentration in terms of grams per liter (g/L) and then calculate the amount needed for a 10 L solution.

A 1 in 500 dextrose solution means that there is 1 part dextrose in 500 parts of the total solution. This can be expressed as a ratio of dextrose to the total solution:

Dextrose:Total solution = 1:500

To convert this ratio into a concentration, we need to calculate the fraction of dextrose in the solution.

Fraction of dextrose = 1 / 500 = 0.002

Now we can calculate the concentration in grams per liter (g/L):

Concentration (g/L) = Fraction of dextrose * 1000

Concentration (g/L) = 0.002 * 1000 = 2 g/L

Now we can calculate the total amount of dextrose needed for a 10 L solution:

Amount of dextrose (g) = Concentration (g/L) * Volume of solution (L)

Amount of dextrose (g) = 2 g/L * 10 L = 20 g

Therefore, 20 grams of dextrose are required to prepare 10 L of a 1 in 500 dextrose solution.

Thus, the correct option is c.

94) What is the amount of Heparin needed to prepare a 750 cc volume of 0.25% solution?
a- 0.250 grams
b- 1.875 grams
c- 18.75 grams
d- 25 grams
e- 75 grams

Solution:

To calculate the amount of Heparin required to prepare a 0.25% solution in a 750 cc volume, we need to determine the concentration in terms of grams per cc (g/cc) and then calculate the amount needed.

A 0.25% Heparin solution means that there are 0.25 grams of Heparin in 100 cc of the solution. This can be expressed as a ratio of Heparin to the total solution:

Heparin:Total solution = 0.25:100

To convert this ratio into a concentration, we can calculate the concentration in grams per cc (g/cc):

Concentration (g/cc) = (0.25/100) g/cc = 0.0025 g/cc

Now we can calculate the total amount of Heparin needed for a 750 cc solution:

Amount of Heparin (g) = Concentration (g/cc) * Volume of solution (cc)

Amount of Heparin (g) = 0.0025 g/cc * 750 cc = 1.875 g

Therefore, 1.875 grams of Heparin are required to prepare a 0.25% solution in a 750 cc volume.

Thus, the correct option is b.

95) To prepare a 5% solution of erythromycin with a volume of 500 cc, how many 500mg tablets of erythromycin are needed?
a- 10 tablets
b- 20 tablets
c- 25 tablets
d- 50 tablets
e- 60 tablets

Solution:

To calculate the number of 500mg tablets of erythromycin needed to prepare a 5% solution with a volume of 500 cc, we need to consider the concentration and dosage of the tablets.

Actually, 1cc = 1mL

First, let's determine the amount of erythromycin required for the solution.

The concentration of the solution is 5%, which means 5 grams of erythromycin are there in 100 mL solution but the required solution is 500 cc, so

grams of erythromycin in 500 cc solution = (5/100) * 500 = 25 grams

So, 5pc 500cc solution of erythromycin will bear 25 grams of erythromycin,

25 grams = 25,000 mg

Since each tablet contains 500mg of erythromycin, the number of tablets required to prepare the solution can be calculated as:

Number of tablets = (Required amount of erythromycin in mg) / (Dosage per tablet in mg)

Number of tablets = 25,000 mg / 500 mg

Number of tablets = 50 tablets

Therefore, you would need a total of 50 tablets of erythromycin (each tablet being 500mg) to prepare a 5% solution with a volume of 500 cc.

Thus, the correct option is d.

96) There is a patient who recently had an accident and experienced a blood loss of about 350 mL. An iron replacement product, iron dextran injection USP whose concentration is 50mg/2 ml is prescribed to compensate for the blood loss.  How much of this injection in milliliters (ml) will be suitable to be given to this patient if he has a hematocrit of 25%?
a- 0.35 ml
b- 3.5 ml
c- 25 mL
d- 35 mL
e- 350 mL

Solution:

To calculate the amount of iron dextran in milliliters (ml) required for a patient with a blood loss of 350mls and a hematocrit of 25%, we need to consider the concentration of the iron dextran solution and the desired amount of iron.

The concentration of iron dextran is given as 50mg/2ml, which means there are 50mg of iron dextran in 2ml of solution.

First, let's calculate the desired amount of iron in milligrams based on the patient's blood loss and hematocrit level.

Desired amount of iron = Blood loss in ml * (Hematocrit / 100)

Desired amount of iron = 350ml * (25 / 100)

Desired amount of iron = 87.5mg

Next, we need to calculate the required volume of iron dextran solution in milliliters based on the desired amount of iron and the concentration of the solution.

Required volume of iron dextran = (Desired amount of iron in mg) * (Volume of iron dextran solution per mg of iron)

Required volume of iron dextran = 87.5mg * (2ml / 50mg)

Required volume of iron dextran = 3.5 ml

Therefore, approximately 3.5 ml of iron dextran solution would be required for a patient with a blood loss of 350mls and a hematocrit of 25%.

Thus, the correct option is b.

97) Following is the formula of a cold cream preparation;

Mineral oil-------------------- 50 parts
Cetyl Esters wax------------- 15 parts
White wax-------------------- 12 parts
Sodium borate--------------- 1 part
Water-------------------------- 22 parts

Calculate the amount in grams of White wax required to make one pound of the cold cream?

a- 12 grams
b- 45.84 grams
c- 54.48 grams
d- 6 grams
e- 24 grams

Solution:

First, lets calculate the total parts of the formula;

total parts = 50 + 15 + 12 + 1 + 22 = 100

Now, to calculate the amount in grams of White wax required to make one pound of the cold cream, we need to convert the weight percentages of the ingredients into actual weights.

Determine the weight of one pound in grams.
One pound is approximately 454 grams

Calculate the weight of White wax.
The White wax is specified as 12 parts in the formula. Since the total number of parts in the formula is 100, we can calculate the weight of White wax as follows:
Weight of White wax = (12 / 100) * weight of one pound

Substituting the weight of one pound (453.592 grams) into the equation:
Weight of White wax = (12 / 100) * 454 grams

Let's perform the calculation:

Weight of White wax = (12 / 100) * 454 grams
Weight of White wax = 54.48 grams

Therefore, 54.48 grams of White wax are required to make one pound of the cold cream.

Thus, the correct option is c.

98) What volume, in milliliters, of alcohol USP is required to make 5 gallons of cough elixir with a 15% v/v ethanol concentration? (Note: 1 gallon = 3,784 mL)
a- 413 mL
b- 717 mL
c- 1,456 mL
d- 2,987 mL
e- 3,124 mL

Solution:

To solve this problem, we need to first convert the ethanol concentration to a decimal value.

15% v/v ethanol concentration means that there are 15 mL of ethanol per 100 mL of cough elixir.

So, the concentration of ethanol in decimal form is 0.15.

Since alcohol USP is equivalent to 95% ethanol. So, the concentration of alcohol USP in decimal form will be 0.95.

Let's also convert gallons into milliliters;

since 1 gallon = 3,784 mL
so, 5 gallons = 18920 mL

Now we can solve the problem by using the formula;

c1v1 = c2v2

where c1 is the initial given concentration,
v1 is the volume of alcohol USP required,
c2 is the desired concentration, and
v2 is the final volume.

In this case, we know:

c1 = 95% (since alcohol USP is 95% ethanol)
c2 = 15% (since the desired concentration is 15% ethanol)
v2 = 5 gallons = 18920 mL

We need to find v1, the volume of alcohol USP required.

Now we can plug in the values and solve for v1:

c1v1 = c2v2

0.95 (v1) = 0.15 (18920)

v1 = 0.15(18920) / 0.95

v1 = 2,987 mL

Therefore, the volume of alcohol USP required to make 5 gallons of cough elixir with a 15% v/v ethanol concentration is 2,987 mL.

Thus, the correct option is d.

99) What quantity (in mL) of 5% w/v aluminum chloride solution should be mixed with a 20% w/v solution to create a 120 mL mixture with a strength of 12%?
a- 40 mL
b- 56 mL
c- 64 mL
d- 74 mL
e- 84 mL

Solution:

We can solve this problem by Alligation method.

The following will be the Alligation diagram.

20%                  7 parts  ........... x mL
12%
5%                    8 parts  ........... y mL
----------
15 parts ........... 120 mL

Since we are to make 120 mL final mixture, so the added parts (15 in total) will be equivalent to 120 ml.

That is,

7 parts + 8 parts = 15 parts

x mL + y mL = 120 mL

Now lets convert the parts into milliliters;

First we will find the required mL of 20% w/v aluminum chloride solution;

7 parts           x mL
---------  =  ------------
15 parts       120 mL

x = 56 mL

Second, we will find the required mL of 5% w/v aluminum chloride solution;

8 parts           y mL
---------  =  ------------
15 parts       120 mL

y = 64 mL

Thus,

56 mL + 64 mL = 120 mL

Therefore, 56 mL of 20% w/v aluminum chloride solution should be mixed with 64 mL of a 5% w/v solution to create a 120 mL mixture with a strength of 12%.

Thus, the correct option is c.

100) Zinc chloride dissociates into three ions (Zn2+, Cl-, Cl-) in weak solutions. 80% of the zinc chloride dissociates in weak solutions. Calculate the dissociation factor for zinc chloride.
a- 0/8
b- 1
c- 1.8
d- 2.2
e- 2.6

Solution:

The formula for dissociation factor is;

dissociation factor = total number of particles after dissociation divided by the original number of particles

ZnCl2 dissociates into Zn2+, Cl-, and Cl- ions in weak solutions which makes it a 3-ion electrolyte.

80% of the ZnCl2 dissociates in weak solutions, so 80% of the particles break apart into ions.

If there are 100 original ZnCl2 particles, then 80 of them dissociate into ions. So there are now 80 Zn2+ ions, 80 Cl- ions, and 80 Cl- ions, for a total of 240 ions.

In addition to the 240 ions, 20 undissociated ZnCl2 particles remain (100 original - 80 dissociated).

So the total number of particles after dissociation are;

240 + 20 = 260 particles

Plugging in the values into the formula;

dissociation factor = 260 particles / 100 original particles = 2.6

So the dissociation factor is 2.6.

In summary:

ZnCl2 (3 ions) dissociates 80%
Original particles: 100
Dissociated particles: 80
Zn2+ ions: 80
Cl- ions: 80 + 80 = 160
Undissociated particles: 20
Total particles after dissociation: 240 + 20 = 260

Dissociation factor = 260 / 100 = 2.6

Thus, the correct answer is e.

WATCH ALSO: 👇👇👇👇👇

www.pharmacymcqs.com