A 6.50 molal solution of $\mathrm{KOH}(\mathrm{aq})$ has a density of $1.89 \mathrm{~g} \mathrm{~cm}^{-3}$. The molarity of the solution is \_\_\_\_ $\mathrm{mol} \mathrm{dm}^{-3}$. (Round off to the nearest Integer). [Atomic masses : $\mathrm{K}=39.0 \mathrm{u}, \mathrm{O}=16.0 \mathrm{u}, \mathrm{H}=1.0 \mathrm{u}$ ]

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Some Basic Concepts of Chemistry

A 6.50 molal solution of

KOH (aq)

has a density of

1.89 g cm^{- 3}

. The molarity of the solution is ____

mol dm^{- 3}

. (Round off to the nearest Integer).
[Atomic masses :

K = 39.0 u, O = 16.0 u, H = 1.0 u

]

Answer: 9

Molality is stated per kilogram of solvent, so take a portion holding 6.50 mol of KOH in 1000 g of water. With a molar mass of

56 g mo l^{- 1}

the solute weighs

6.50 \times 56 = 364

g, and the solution weighs

364 + 1000 = 1364

g.
Its volume follows from the density:

V = \frac{1364}{1.89} = 721.7 c m^{3} = 0.7217 d m^{3}

M = \frac{6.50}{0.7217} = 9.01

Rounded to the nearest integer, the molarity of the solution is 9 mol dm

^{- 3}

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Integer answer