Question

How many gram of $NaOH$ will be required to neutralize $12.2$ g benzoic acid?

• A. $12.2$ g
• B. $16$ g
• C. $40$ g
• D. $4$ g

Answer 1

Answer: (D)

$4$ g

Benzoic acid $⟶$ (Refer to Image)

Number of $H=6=6g$

$C=7=12\times 7g$

$O=2=16\times 2g$

$122g$ weight of molecule

Number of moles in $12.2g=\frac{12.2}{122}=0.1$ mole

Moles are always equal in reaction

So, $0.1$ mole=$\frac{Weight}{Molarmass}=\frac{x}{40}$

$\Rightarrow x=4g$