Molality of $0.8 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ solution (density $1.06 \mathrm{~g} \mathrm{~cm}^{-3}$ ) is ____ $\times 10^{-3} \mathrm{~m}$.

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Some Basic Concepts of Chemistry

Molality of

0.8 M H_{2} SO_{4}

solution (density

1.06 g cm^{- 3}

) is ____

\times 1 0^{- 3} m

Answer: 815

Molality relates to molarity through the density of the solution and the molar mass of the solute:

m = \frac{M \times 1000}{d _{sol} \times 1000 - M \times M _{solute}}

With

M = 0.8

d_{sol} = 1.06 g c m^{- 3} = 1060 g L^{- 1}

, and molar mass of

H_{2} S O_{4} = 98 g mo l^{- 1}

m = \frac{0.8 \times 1000}{1060 - 0.8 \times 98} = \frac{800}{981.6} = 0.815 mol k g^{- 1}

The molality is

815 \times 1 0^{- 3} m

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Integer answer