The molarity of the solution prepared by dissolving 6.3 g of oxalic acid $\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right)$ in 250 mL of water in $\mathrm{mol} \mathrm{L}^{-1}$ is $x \times 10^{-2}$. The value of $x$ is ______. (Nearest integer) [Atomic mass : H : 1.0, C : 12.0, O : 16.0]

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Some Basic Concepts of Chemistry

The molarity of the solution prepared by dissolving 6.3 g of oxalic acid

(H_{2} C_{2} O_{4} \cdot 2 H_{2} O)

in 250 mL of water in

mol L^{- 1}

x \times 1 0^{- 2}

. The value of

x

is ______. (Nearest integer)
[Atomic mass : H : 1.0, C : 12.0, O : 16.0]

Answer: 20

The molar mass of

H_{2} C_{2} O_{4} \cdot 2 H_{2} O

2 (1) + 2 (12) + 4 (16) + 2 (18) = 126 g mo l^{- 1}

. Dissolving

6.3 g

gives

\frac{6.3}{126} = 0.05 mol

, and in

0.250 L

the molarity is

\frac{0.05}{0.250} = 0.20 mol L^{- 1} = 20 \times 1 0^{- 2} mol L^{- 1}

. The value of

x

20

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Integer answer