Question

10.6 g of a substance of molecular weight 106 was dissolved in 100 ml. 10 ml of this solution was pipetted out into a 1000 ml flask and made up to the mark with distilled water. The molarity of the resulting solution is:

• A. $1M$
• B. ${10}^{-2}M$
• C. ${10}^{-3}M$
• D. ${10}^{-4}M$

Answer 1

The correct option is B ${10}^{-2}M$

Initial molarity-

$=\frac{10.6}{106}\times \frac{1000}{V\left(ml\right)}$

$=\frac{10.6}{106}\times \frac{1000}{100}$ $=1M$

Now, $M_1{V}_1=M_2{V}_2$

$1\times 10=M_2\times 1000$

$M_2=\frac{10}{1000}$

$M_2={10}^{-2}M$

So, the correct option is $B$.