Question

0.84 g of an acid (molecular weight 150) was dissolved in water and made up to 100 ml 25 ml of this solution required 28 ml of deci normal NaOH solution for neutralization.The equivalent weight and basicity of the acid is:

• A. 75
• B. 150.1
• C. 79.4
• D. 150.2

Answer 1

Answer: (A)

75

No. of moles of acid $=\frac{0.84g}{150g{mol}^{-1}}=5.6\times {10}^{-3}$moles

No. of moles of $NaOH$ needed to neutralize $=0.1\times 28\times {10}^{-3}$ moles $=2.8\times {10}^{-3}$ moles

now no. of moles of acid $=$ (no. of moles of $NaOH$) $\times$ $n$

Where $n=$ basicity

$n=\frac{5.6\times {10}^{-3}}{2.8\times {10}^{-3}}=2$

$\therefore$ Equivalent weight of the acid $=\frac{150}{2}=75.0$