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

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Some Basic Concepts of Chemistry

A 100 mL solution was made by adding 1.43 g of

Na_{2} CO_{3} . \times 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

)

Answer: 10

0.1 N

solution in

100 mL

contains

0.1 \times 0.1 = 0.01

equivalents, and since

N a_{2} C O_{3}

has an

n

-factor of

2

, that is

0.005 mol

N a_{2} C O_{3} \cdot x H_{2} O

. Its molar mass is therefore

\frac{1.43}{0.005} = 286 g mo l^{- 1}

, and setting

106 + 18 x = 286

gives

18 x = 180

, so

x = 10

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Integer answer