Math Practice (copy}

Make a 30% w/w NaCl aqueous solution:

Add 30g NaCl to 70g water

Make 100g of 30% NaCl aqueous solution:

Add 30g NaCl to 70g water (total 100g)

Prepare 100mL of a 0.5% NaCl solution:

Add H20 to 0.5g NaCl up to 100mL

Prepare a 25% H2SO4 solution:

Add 25mL of H2SO4 to 75mL H2O

How much 95% alcohol is required to make 200 mL of 5% alcohol?

10.5 mL

Prepare 500mL of 0.5% NaOH (how many grams of NaOH is needed?)

2.5g

A solution contains 45g of solute in 240 mL of solution. What is the % concentration?

18.75%

Make 500mL of 2M NaCl (GMW = 58.5) (How many grams of NaCl needs to be diluted in 500mL?)

58.5g

What is the molarity of a solution in which there are 25g of Na2SO4 in 500mL of solution? GMW = 142

0.35M

Absorbance of the unknown = 0.252

Standard absorbance = 0.640

Standard concentration = 200 mg/dL

Concentration of the unknown = (?)

79 mg/dL

Express 3/15 as a percentage

20%

Express 0.7 as a percentage

0.7%

If there are 5 g of NaCl in 100 mL of solution, how many mL of the same concentration would 10 g of NaCl make?

200 mL

How many grams of substance would be required to make 100 mL of solution if 5 g will make 250 mL?

2

If 300 uL of saline is added to 30 uL of serum, what is the dilution ratio of serum to saline?

1:10 serum:saline

How much of a 1/50 dilution can be made with 500 uL of urine?

25,000 uL solution

A 10% solution is to be diluted with water using a series of four dilutions. The dilution in tube 1 is 1/5, followed by a three fold dilution series The dilution in the fourth tube would be?

1/135

If 100 uL of solution serum in saline contains 90 uL of saline, what is the ratio for the solution?

1:9 serum to saline

What is the resulting concentration of a solution of 30mL of a 4.5N is diluted up to 200 mL?

0.675 N

A white cell count is done on a blood sample that has been diluted 1 to 25. The number of cells counted in the sample is 565 cells per milliliter cubed. What number should be reported for the undiluted sample?

14,125 cells/mL^3

Change 6.25 x 10^-5 to a decimal:

0.0000625

Change 6.25 x 10^5 to a decimal

625,000

Change 3.5 x 10^1 to a decimal

35

Convert 653 mg to ng

6.53 x 10^8 ng

Convert 300 mL to L

3 x 10^-1

Convert 0.065 mm to nm

6.5 x 10^4 nm

An automated CK assay gives a reading that is above the limits of linearity. A dilution of the serum sample is made by adding 1 mL of serum to 9 mL of water. The instrument now reads 350 U/L. The correct report on the undiluted serum should be:

a. 2,850 U/L

b. 3,150 U/L

c. 3.500 U/L

d. 3,850 U/L

c. 3.500 U/L

How many mL of red blood cells are to be used to make 25 mL of a 4% red cell suspension?

a. 0.25 mL

b. 0.5 mL

c. 1 mL

d. 2 mL

c. 1 mL

The volume of 25% stock solution needed to prepare 100 mL of 5% working solution is:

a. 1.25 mL

b. 5 mL

c. 20 mL

d. 50 mL

c. 20 mL

How many grams of sodium chloride are needed to prepare 1 L of 0.9% normal saline?

a. 0.9

b. 1.8

c. 9.0

d. 18.0

c. 9.0

The following data were calculated on a series of 30 determinations of serum uric acid control: mean = 5.8 mg/dL, 1 standard deviation = 0.15 mg/dL. If confidence limits are set at ±2 standard deviations, which of the following represents the allowable limits for the control?

a. 5.65 - 5.95 mg/dL

b. 5.35 - 6.25 mg/dL

c. 5.50 - 6.10 mg/dL

d. 5.70 - 5.90 mg/dL

c. 5.50 - 6.10 mg/dL

The mean value of a series of hemoglobin controls was found to be 15.2 g/dL, and the standard deviation was calculated at 0.20. Acceptable control range for the laboratory is ±2 standard deviations. Which of the following represents the allowable limits for the control?

a. 14.5 - 15.5 g/dL

b. 15.0 - 15.4 g/dL

c. 15.2 - 15.6 g/dL

d. 14.8 - 15.6 g/dL

d. 14.8 - 15.6 g/dL

A mean value of 100 and a standard deviation of 1.8 mg/dL were obtained from a set of glucose measurements on a control solution. The 95% confidence interval in mg/dL would be:

a. 94.6 - 105.4

b. 96.4 - 103.6

c. 97.3 - 102.7

d. 98.2 - 101.8

b. 96.4 - 103.6

When a 0.25 mL is diluted to 20 mL, the resulting dilution is:

a. 1:20

b. 1:40

c. 1:60

d. 1:80

d. 1:80

A serum glucose sample was too high to read, so a 1:5 dilution using saline (dilution A) was made. Dilution A was tested and was again too high to read. A further 1:2 dilution was made using saline (dilution B). To calculate the result, the dilution B value must be multiplied by:

a. 5

b. 8

c. 10

d. 20

c. 10

If 0.5 mL of a 1:300 dilution contains 1 antigenic unit, 2 antigenic units would be contained in 0.5 mL of a dilution of:

a. 1:150

b. 1:450

c. 1:500

d. 1:600

a. 1:150

A colorimetric method calls for the use of 0.1 mL of serum, 5 mL of reagent and 4.9 mL of water. What is the dilution of the serum in the final solution?

a. 1:5

b. 1:10

c. 1:50

d. 1:100

d. 1:100

4 mL of water are added to 1 mL of serum. This represents which of the following serum dilutions?

a. 1:3

b. 1:4

c. 1:5

d. 1:6

c. 1:5

A solution contains 20 g of solute dissolved in 0.5 L of water. What is the percentage of this solution?

a. 2%

b. 4%

c. 6%

d. 8%

b. 4%

How many grams of sulfosalicylic acid (MW = 254) are required to prepare 1 L of a 3% (w/v) solution?

a. 3

b. 30

c. 254

d. 300

b. 30

How many mL of a 3% solution can be made if 6 grams of solute are available?

a. 100 mL

b. 200 mL

c. 400 mL

d. 600 mL

b. 200 mL

The following 5 sodium control values in unit (mEq/L) were obtained:

140, 135, 138, 140, 142

Calculate the coefficient of variation.

a. 1.9%

b. 2.7%

c. 5.6%

d. 6.1%

a. 1.9%

Given the following values:

100, 120, 150, 140, 130

What is the mean?

a. 100

b. 128

c. 130

d. 640

b. 128

A cholesterol QC chart has the following data for the normal control:

mean = 137 mg/dL

#x = 1,918 mg/dL

2 SD = 6 mg/dL

N = 14

The coefficient of variation for this control is:

a. 1.14%

b. 2.19%

c. 4.38%

d. 9.49%

b. 2.19%

The sodium content (in grams) in 100 grams of NaCl is approximately:

(atomic weights: Na = 23.0, Cl = 35.5)

a. 10

b. 20

c. 40

d. 60

c. 40

80 grams of NaOH (MW = 40) are how many moles?

a. 1

b. 2

c. 3

d. 4

b. 2

How many mL of 0.25 N NaOH are needed to make 100 mL of a 0.05 N solution of NaOH?

a 5 mL

b 10 mL

c 15 mL

d 20 mL

d 20 mL