Question

$250\, mL$ of a $Na_2CO_3$ solution contains $2.65 \,g$ of $ Na_2CO_3 \cdot 10\,mL $ of this solution is added to $ x \,mL $ of water to obtain $ 0.001\, M $ $ Na_2CO_3 $ solution. The value of $ x $ is (molecular weight of $ Na_2CO_3 = 106 $ )

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

Answer 1

Answer: (B)

Given, volume of $Na _{2} CO _{3}$ solution $=250\, mL$

and, mass of $Na _{2} CO _{3}$ solution $=2.65\, g$

molecular mass of $Na _{2} CO _{3}=106 \,g$

$\therefore $ Eq wt of $Na _{2} CO _{3}=\frac{106}{2}=53$

$ \because w =\frac{E N V}{1000} $

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

$ N =0.2 $

$\because \,\,10 \,mL$ of this solution is added to $x mL$ of water to obtain $0.001 M Na _{2} CO _{3}$ solution, therefore

volume of solution taken, $V_{1}=10 \,mL$

normality of solution, $N_{1}=0.2\, N$

volume of solution after addition of water

$V_{2}=10+x$

Normality of solution,

$N_{2}=0.001 \,M =0.002 \,N$

$\left(\because\right.$ For $\left.Na _{2} CO _{3}, N=2\, M \right)$

From, $N_{1} V_{1}=N_{2} V_{2}$

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

$x=990\, mL$