Question

The Molarity of $200ml$ of $HCl$ solution which can neutralise $1.6g$ of anhydrous ${Na}_2{CO}_3$ is:

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

Answer 1

Answer: (B)

$0.1M$

$2HCl+N{a}_{2}C{O}_3⟶2NaCl+H_{2}C{O}_3$

$N_1{V}_1=N_2{V}_2$

$\Rightarrow M\times 200=\frac{1.06}{106}\times 1000\times 2$

$\Rightarrow M_1=\frac{10.6}{106}=0.1M$

$M_{1}V=\frac{W}{GEW}\times 1000$

$\therefore$ Given question $1.6g$ of $N{a}_{2}C{O}_3$ should be replaced by $1.06g$ then $weg$ at $M=0.1M$.