Question

$250mL$ of a ${Na}_2{CO}_3$ solution contains $2.65g$ of ${Na}_2{CO}_3\cdot 10mL$ of this solution is added to $x$ $mL$ of water to obtain $0.001M$ ${Na}_2{CO}_3$ solution. The value of $x$ is:

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

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

Answer 1

The correct option is B $990mL$

Given, volume of ${Na}_2{CO}_3$ soluton $=250mL$
and mass of ${Na}_2{CO}_3$ soluton $=2.65g$

Molecular mass of ${Na}_2{CO}_3$ soluton $=106g$
$\therefore$ Eq.wt of ${Na}_2{CO}_3=\frac{106}{2}=53$
$\because w=\frac{ENV}{1000}$

$2.65=\frac{53\times N\times 250}{1000}$

$N=0.2$

$\because$ $10mL$ of this solution is added to $x$ $mL$ of water to obtain $0.001M$ ${Na}_2{CO}_3$ solution,

therefore
volume of solution taken, $V_1=10mL$
normality of solution, $N_1=0.2N$

Volume of solution after the addition of water
$V_2=10+x$

Normality of solution,
$N_2=0.001M=0.002N$ (For ${Na}_2{CO}_3,N=2M$)

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

$10\times 0.2=(10+x)\times 0.002$

$x=990mL$