Molarity and normality

8 ASCP MLS practice questions on Molarity and normality, part of Laboratory Operations. The first 8 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Hard

   A technologist must standardize a sodium hydroxide titrant against a primary standard of oxalic acid dihydrate (H2C2O4·2H2O, formula weight 126.07; two acidic hydrogens, both titrated to the phenolphthalein end point). The procedure calls for 1.000 L of 0.1000 N oxalic acid. Using the formula weight, she dissolves 12.61 g of the dihydrate in water to exactly 1.000 L and labels the flask 0.1000 N. She then titrates 20.00 mL of this solution, which needs 40.00 mL of the NaOH to reach the end point. What is the true normality of the NaOH?

   1. A0.4000 N
   2. B0.0500 N
   3. C0.1000 N
   4. D0.2000 N

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   Answer: C · 0.1000 N

   Oxalic acid gives 2 equivalents per mole, so its equivalent weight is 126.07 ÷ 2 = 63.04 g, and 0.1000 N needs 6.304 g per liter. The 12.61 g she weighed is 0.1000 mol, which makes the solution 0.1000 M but 0.2000 N, twice the labeled normality. At the end point, equivalents of acid equal equivalents of base: N(NaOH) × 40.00 mL = 0.2000 N × 20.00 mL, so N(NaOH) = 0.1000 N. Calculated from the wrong label, the NaOH would be called 0.0500 N and every result based on it would be off by a factor of 2, so the flask must be relabeled 0.2000 N or remade.

   • A: 0.4000 N uses the true 0.2000 N but swaps the two volumes.
   • B: 0.0500 N uses the label of 0.1000 N, but the solution is really 0.2000 N.
   • D: 0.2000 N uses the label value with the two volumes swapped.

2. Question 2 · Hard

   Concentrated hydrochloric acid is 37% HCl by weight, with a specific gravity of 1.19 (HCl molecular weight 36.5). About how much is needed to make 1.0 L of 1.0 M HCl?

   1. A99 mL
   2. B117 mL
   3. C31 mL
   4. D83 mL

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   Answer: D · 83 mL

   One liter of 1.0 M HCl needs 36.5 g of HCl. The concentrate is only 37% HCl, so 36.5 ÷ 0.37 = 98.6 g of concentrate is needed. Each mL weighs 1.19 g, so the volume is 98.6 ÷ 1.19 = 83 mL (the concentrate is about 12 M). Add the acid to part of the water, never water to acid, then make up to volume.

   • A: 99 mL treats grams of concentrate as milliliters, ignoring the specific gravity.
   • B: 117 mL multiplies by the specific gravity instead of dividing by it.
   • C: 31 mL ignores that the concentrate is only 37% HCl.

3. Question 3 · Medium

   Normal saline is 0.9% (w/v) sodium chloride (molecular weight 58.5). Its molarity is about:

   1. A1.54 M
   2. B0.0154 M
   3. C0.154 M
   4. D0.308 M

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   Answer: C · 0.154 M

   A 0.9% (w/v) solution holds 0.9 g per 100 mL, which is 9 g per liter. Molarity = 9 g/L ÷ 58.5 g/mol = 0.154 mol/L, or 154 mmol/L each of sodium and chloride. Because NaCl splits into two ions, its osmolarity is about 308 mOsm/L, close to that of plasma. Converting % (w/v) to g/L first (multiply by 10) avoids decimal errors.

   • A: 1.54 M treats 0.9% as 90 g per liter.
   • B: 0.0154 M treats 0.9% as 0.9 g per liter instead of 9 g.
   • D: 0.308 M counts both ions, which gives osmolarity, not molarity.

4. Question 4 · Hard

   A reagent procedure calls for 500 mL of 0.100 M copper sulfate and lists the formula weight of anhydrous CuSO4 (159.6). Only the pentahydrate, CuSO4·5H2O (formula weight 249.7), is available. How much of it should be weighed?

   1. A12.5 g
   2. B5.10 g
   3. C25.0 g
   4. D7.98 g

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   Answer: A · 12.5 g

   The solution needs 0.100 mol/L × 0.500 L = 0.0500 mol of copper sulfate, whatever form it is weighed in. Each mole of the pentahydrate weighs 249.7 g because it includes five waters, so 0.0500 × 249.7 = 12.5 g. Weighing 7.98 g of the hydrate would give only about 0.032 mol, a solution about 36% too weak. Always use the formula weight of the form actually on the shelf.

   • B: 5.10 g applies the formula-weight ratio upside down.
   • C: 25.0 g is the amount for a full liter, not 500 mL.
   • D: 7.98 g is correct only for the anhydrous salt; the hydrate needs more.

5. Question 5 · Hard

   In a standardization, 25.0 mL of sodium hydroxide is exactly neutralized by 15.0 mL of 0.100 M sulfuric acid. The molarity of the sodium hydroxide is:

   1. A0.120 M
   2. B0.167 M
   3. C0.333 M
   4. D0.060 M

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   Answer: A · 0.120 M

   Sulfuric acid gives two hydrogen ions, so 0.100 M H2SO4 is 0.200 N. At the endpoint, N1 × V1 = N2 × V2: 0.200 × 15.0 = N × 25.0, so N = 0.120. NaOH has one hydroxide per molecule, so 0.120 N equals 0.120 M. Solid NaOH absorbs water and carbon dioxide, so its solutions are standardized by titration rather than trusted by weight alone.

   • B: 0.167 M swaps the two volumes and ignores the acid's two protons.
   • C: 0.333 M swaps the two volumes in the equation.
   • D: 0.060 M treats the acid as giving one hydrogen ion instead of two.

6. Question 6 · Easy

   A 1.0 molar (1.0 M) solution contains:

   1. AOne equivalent of solute per liter
   2. BOne mole of solute per kg of solvent
   3. COne mole of solute per liter of solution
   4. DOne gram of solute per 100 mL

   Answer and explanation in LearnFree account

7. Question 7 · Medium

   A patient's serum calcium is 9.0 mg/dL. Calcium has an atomic weight of 40 and a valence of 2. What is this result in mEq/L?

   1. A0.45 mEq/L
   2. B1.13 mEq/L
   3. C4.5 mEq/L
   4. D2.25 mEq/L

   Answer and explanation in LearnFree account

8. Question 8 · Medium

   How many grams of anhydrous sodium carbonate (Na2CO3, molecular weight 106.0; each mole neutralizes 2 mol of acid) are needed to make 200 mL of a 0.100 N solution?

   1. A4.24 g
   2. B1.06 g
   3. C2.12 g
   4. D5.30 g

   Answer and explanation in LearnFree account