The mole fraction of $C H_{3} O H$ in an aqueous solution is 0.02 and its density is $0.994 g c m^{- 3}$ . Determine its molarity and molality.

Open in App

Solution

Let $x$ mole of $C H_{3} O H$ and y mole of water be present in solution.
Mole fraction of $C H_{3} O H = \frac{x}{x + y} = 0.02$
So,
$\frac{y}{x} = 49$ or $\frac{x}{y} = \frac{1}{49}$

Molality $= \frac{x}{18 \times y} \times 1000 = \frac{1000}{18 \times 49} = 1.13 m$

Volume of solution $= \frac{T o t a l m a s s}{d e n s i t y} = \frac{32 x + 18 y}{0.994} m L$

$= \frac{32 x + 18 y}{0.994 \times 1000} l i t r e = \frac{32 x + 18 y}{994} l i t r e$

Molarity $= \frac{x}{32 x + 18 y} \times 994$

$= \frac{994}{32 + 18 \times y / x} = \frac{994}{32 + 18 \times 49} = 1.0875 M$

Suggest Corrections

Similar questions

Q. Determine mole fraction of

C H_{3} O H

in an aqueous solution of

C H_{3} O H

whose molality is 2.00 m.

Q. An aqueous solution of

N a O H

having density

1.1

k g / {d m}^{3}

contains

0.02

mole fraction of

N a O H

. The molality and molarity of

N a O H

solution respectively, are :

Q. Determine mole fraction of

C H_{3} O H

in an aqueous solution of

C H_{3} O H

whose molality is 2.00 m.

Molarity of H2SO4 solution is 0.8M and its density is 1.06g/cm-3 What will be the concentration of the solution in terms of molality and determine the mole fraction of the solution

Q. An aqueous solution of sodium chloride is marked

10

(\frac{w}{w})

on the bottle. The density of the solution is

1.071 g m l^{- 1} .

What is its molality and molarity? Also, What is the mole fraction of each component in the solution?

Join BYJU'S Learning Program

The mole fraction of CH3OH in an aqueous solution is 0.02 and its density is 0.994gcm−3. Determine its molarity and molality.

The mole fraction of $C H_{3} O H$ in an aqueous solution is 0.02 and its density is $0.994 g c m^{- 3}$ . Determine its molarity and molality.