Question

0.45 g of an acid of mol.wt. 90 was neutralised by 20 mL of 0.54 N caustic potash (KOH). The basicity of acid is:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

Weight of acid $=0.45g$

Moles $=\frac{0.45}{90}=\frac{1}{2}\times {10}^{-2}=5$ millimoles

$N_1{V}_1=N_2{V}_2$

$(5\times {10}^{-3})\times basicity=20\times 0.54\times {10}^{-3}$

$basicity\approx 2$

Hence, option B is correct.