JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Some Basic Concepts of Chemistry

6.023 \times 1 0^{22}

molecules are present in 10 g of a substance 'x'. The molarity of a solution containing 5 g of substance 'x' in 2 L solution is

____

\times 1 0^{- 3}

Answer: 25

Solution

First fix the molar mass of x. Since

6.023 \times 1 0^{22}

molecules is

\frac{6.023 \times 1 0 ^{22}}{6.023 \times 1 0 ^{23}} = 0.1 mol

and this weighs

10 g

, the molar mass is

100 g mol^{- 1}

. Then

5 g

of x is

\frac{5}{100} = 0.05 mol

, and in

2 L

the molarity is

\frac{0.05}{2} = 0.025 M

, that is

25 \times 1 0^{- 3} M

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Integer answer