Molality of 0.8\ M\ H 2SO 4 solution (density 1.06\ g\ cm -3 ) is \ \ \ \ \ × 10 -3 \ m .

Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2024 — 29 January, Shift 2 — Question 78

Molality of $0.8\ \mathrm{M\ H_2SO_4}$ solution (density $1.06\ \mathrm{g\ cm^{-3}}$ ) is $\_\_\_\_\_$ $\times 10^{-3}\ \mathrm{m}$ .

Numerical answer — enter this value.

Take $1\ L$ solution.

Moles of $\mathrm{H_2SO_4}$ $= \mathrm{0.8\ mol}$

Mass of solution $= \mathrm{1.06 \times 1000 = 1060\ g}$

Mass of solute $= \mathrm{0.8 \times 98 = 78.4\ g}$

Mass of solvent $= \mathrm{1060 - 78.4 = 981.6\ g = 0.9816\ kg}$

Molality, $\mathrm{m = \frac{0.8}{0.9816} \approx 0.815}$

\mathrm{m = 815 \times 10^{-3}}

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.