Question

5.85 g of NaCl is present in a solution. If normality of solution is 0.5 N, then volume of solution (in mL) is

• A. 100 mL
• B. 200 mL
• C. 300 mL
• D. 400 mL

Answer 1

Answer: (B)

200 mL

Answer 2

1. Molar mass of NaCl = 58.5 g/mol.
2. n-factor for NaCl = 1.
3. Equivalent weight of NaCl = 58.5 g/equivalent.
4. Gram equivalents of NaCl = 5.85 g / 58.5 g/equivalent = 0.1 equivalent.
5. Normality (N) = Gram equivalents / Volume (L).
6. 0.5 N = 0.1 equivalent / Volume (L).
7. Volume (L) = 0.1 / 0.5 = 0.2 L.
8. Volume (mL) = 0.2 L * 1000 mL/L = 200 mL.