Question

The Molality of $200ml$ of $HCl$ solution which can neutralizes $10.6g$. of anhydrous $N{a}_{2}C{O}_3$ is:

• A. $0.1M$
• B. $1M$
• C. $0.6M$
• D. $0.75M$

Answer 1

Answer: (B)

$1M$

Moles of anhydrous $N{a}_{2}C{O}_3=\frac{10.6}{106}=0.1$ moles

$2$ moles of $HCl$ are required for $1$ mole of $N{a}_{2}C{O}_3$

$\therefore$ Moles of $HCl=0.2$ (for $0.1$ mole of $N{a}_{2}C{O}_3$)

$\therefore$ Molarity= $\frac{molesofHCl}{Volume}=\frac{0.2}{0.2}=1$ M