Question

How much of $NaOH$ is required to neutralise $1500{cm}^3$ of $0.1N$ $HCl$ ($Na=23$)?

• A. $40g$
• B. $4g$
• C. $6g$
• D. $60g$

Answer 1

The correct option is C $6g$

$N=\frac{W\left(gm\right)\times 1000}{V\times Eq.wt.}$

$1500ml$ of $0.1NHCl=150ml\left(N\right)$

$1=\frac{W\left(gm\right)\times 1000}{150\times 40},W\left(gm\right)=\frac{150\times 40}{1000}=6gm.$