Question

The mass of 60% HCl by mass required for the neutralisation of 10 L of 0.1 M NaOH is

• A. 60.8 g
• B. 21.9 g
• C. 100 g
• D. 219 g

Answer 1

Answer: (A)

60.8 g

Answer 2

$\underset{\text{36.5 g}}{ {Hcl}}+ \underset{\text{40 g}}{ {NaOH}}\to Nacl+H_{2}O$ Amount of $NaOH$ to be neutralised $=0.1 \times 10\times40=40\,g$ $\therefore$ $HCl$ required $= 36. 5\, g$ $60\%$ HCl required $=\frac{36.5}{60}\times100$ $=60.8\,g$