Question

The molarity of the solution containing 2.8% (m/v) solution of KOH is (Given atomic mass of K = 39 u)

• A. 0.1 M
• B. 0.5 M
• C. 0.2 M
• D. 1M

Answer 1

The correct option is B 0.5 M

Weight of KOH = 2.8 g
Volume of solution = 100 mL
Molar mass of KOH = $39+16+1=56gmo{l}^{-1}$
Molarity of the solution $M=\frac{2.8\times 1000}{56\times 100}=\frac{28}{56}=\frac{1}{2}=0.5M$