Question

0.90 gm of the acid (molecular weight 90 gm) is neutralized by 40 mL of 0.5N NaOH. What is the basicity of the acid?

• A. $1$
• B. $2$
• C. $3$
• D. $44$

Answer 1

The correct option is B $2$

Given weight of acid $=0.90gm$

molecular $w=90$

neutralized by 40 ml, $0.5\text{M}$

$E_{eq}=\frac{40\times 0.5}{1000}=\underset{–}{2}\times {10}^{-3}$

mole of $\mathrm{acid}used=\frac{\omega t}{m\omega }=\frac{0.5}{90}=\frac{1}{1000}={10}^{-3}$

Hence =$\frac{2\times {10}^{-3}}{{10}^{-3}}$

$M,V_1=M_2{V}_2$

${10}^{-3}\times V_1=\frac{40}{1000}\times 0.5$

$V_1=2$

Correct Answer $-\underset{–}{B}$