Question

$250ml$ of a sodium carbonate solution contains $2.65$ grams of ${Na}_2{CO}_3$. $10ml$ of this solution is added to ${}^{\prime }{x}^{\prime }ml$ of water to obtain $0.001M$ ${Na}_2{CO}_3$ solution.

What is the value of ${}^{\prime }{x}^{\prime }$ in $ml$?

[Molecular weight of ${Na}_2{CO}_3=106]$

• A. $1000$
• B. $990$
• C. $9990$
• D. $90$

Answer 1

The correct option is B $990$

$Molarity=\frac{weight}{molecularweight}\times \frac{1000}{V\left(ml\right)}$
$M_1=\frac{2\cdot 65}{106}\times \frac{1000}{250}$
$M_1=0.1M$
Now, $M_1{V}_1=M_2{V}_2$
$10\times 0\cdot 1=0\cdot 001\times V_2$
$V_2=1000ml$
$Wateradded=V_2-V_1=1000-10=990ml$