What volume of 14 N H2SO4 is required to prepare 250 mL of 3.2 M H2SO4 (MW 98.08 g/mol)?

• A. 114 mL
• B. 1.82 mL
• C. 1.75 mL
• D. 7 mL

Answer: (A)

Think of how dilution works with a diprotic acid. Normality for H2SO4 counts two replaceable hydrogens, so a 14 N solution corresponds to a molarity of 14 N ÷ 2 = 7 M.

We need 250 mL (0.250 L) of 3.2 M H2SO4. The amount of H2SO4 required in moles is 3.2 mol/L × 0.250 L = 0.80 mol.

From the stock solution at 7 M, the volume needed to provide 0.80 mol is V = n/M = 0.80 mol / 7 mol/L ≈ 0.114 L = 114 mL.

Using the dilution relation M1V1 = M2V2 also gives 7 × V1 = 3.2 × 0.250, yielding V1 ≈ 0.114 L.

Thus the required volume is about 114 mL. The other options are far too small to supply 0.80 mol from a 7 M stock.