Question

What is the volume of $0.1 \, N-HCl $ required to react completely with $1.0 \,g$ of pure calcium carbonate?

• A. $100 \,cm ^{3}$
• B. $150 \,cm ^{3}$
• C. $250 \,cm ^{3}$
• D. $200 \,cm ^{3}$

Answer 1

Answer: (D)

Use the following formula to find $V$ of $HCl$ used
$V=\frac{w \times 1000}{\text { equivalent weight } \times N}$
Given, $w=1.0 g N =0.1$
$\underset{40+16 \times 3+12=100}{ CaCO _{3}}+2 HCl \rightarrow CaCl _{2}+ H _{2} O$
$\therefore 1$ mole of $CaCO _{3}$ reacts with $2$ moles of $HCl$
$\therefore $ Equivalent weight $=\frac{100}{2}=50$
$\therefore V=\frac{1 \times 1000}{50 \times 0.1}$
$\approx 200\, cm ^{3}$