Question

What weight of $80\%$ pure $NaOH$ required to neutralize $10\%$ pure $100ml,1M{H}_{2}S{O}_4$ solution?
[Atomic mass of $Na=23,H=1,O=16,S=32$]

• A. $1gm$
• B. $0.2gm$
• C. $4gm$
• D. $0.5gm$

Answer 1

The correct option is A $1gm$

Given : $10\%$ pure $100ml$, $1M{H}_{2}S{O}_4$ sol

: $80\%$ pure $NaOH$

To find : at of $NaOH$ required for neutralization A/C

As $10\%$ pure $100ml$, $1M.H_{2}S{O}_4$

$\therefore$ mole of $H_{2}S{O}_4=\frac{10}{100}\times \frac{100}{1000}\times 1=0.01$ mole $H_{2}S{O}_4$

Now $2$ mole $NaOH$ required for neutralization of $1$ mole $H_{2}S{O}_4$.

$\therefore$ mole of $NaOH$ required $=2\times 0.01$

$=0.02moleNaOH$

$\therefore$ at of $NaOH$ $=0.02\times 40=0.8gmNaOH$

as it is $80\%$ pure

$\therefore$ mass required $\times \frac{80}{100}=0.8gm$

$\therefore NaOH$ required $\times \frac{0.8}{80}=1gm$