Q23 / 75 · JEE Main 2020 (04 Sep Shift 2) · Chemistry

A $100 mL$ solution was made by adding $1.43 g$ of ${Na}_{2} {CO}_{3} . {xH}_{2} O .$ The normality of the solution is $0.1 N$ . The value of $x$ is _______
(The atomic mass of $Na$ is $23 g / mol$ )

Correct answer: 10

Solution

Equivalent of solute $= 0.1 \times 0.1$

Mole of solute $({Na}_{2} {CO}_{3}, {xH}_{2} O) = [0.1 \times 0.1] \frac{1}{2}$

Mass of ${Na}_{2} {CO}_{3} . {xH}_{2} O = [0.1 \times 0.1] \frac{1}{2} \times [106 + 18 x] = 1.43$

$\Rightarrow [106 + 18 x = 286$

$18 x = 180$

$x = 10$

Hence, the correct answer is 10.