Question

The molarity of the solution containing $2.8\%$ (w/V) solution of $KOH$ is:
(Given atomic mass of $K=39u$)

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