Percent by mass of a solute (molar mass = 28 g) in its aqueous solution is 28. Calculate the mole fraction (X) and molality (m) of the solute in the solution.

X = 0.2, m = 10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

X = 0.2, m = $\frac{125}{9}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

X = 0.8, m = $\frac{125}{9}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

X = 0.8, m = 10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
X = 0.2, m = $\frac{125}{9}$

Mol. of solute in 100 g solution $= \frac{28}{28} = 1$
Mol. of water in 100 g solution $= \frac{100 - 28}{18} = 4$
Mol. fraction of solute $= \frac{1}{1 + 4} = 0.2;$
Molality $= \frac{1 \times 1000}{72} = \frac{125}{9}$

Suggest Corrections

Similar questions

Percent by mass of a solute (molar mass = 28 g) in its aqueous solution is 28. Calculate the mole fraction (X) and molality (m) of the solute in the solution.

Q. Molality of a solution in aqueous medium is

0.8

. Calculate its mole fraction and the percentage by mass of solute if molar mass of solute is

60.

Q. The molality of a solution in water is 10 m. Calculate mole fraction of the solute given the molar mass of the solute is 38.8

g m o l^{- 1}

and molar mass of the solvent = 25

g m o l^{- 1}

Q. Calculate the molality of a 1.5 M aqueous salt solution in a solvent X, if the mole fraction of the salt in the solution is 0.25.
(Molar mass of solvent = 40 g/mol)

Join BYJU'S Learning Program

Percent by mass of a solute (molar mass = 28 g) in its aqueous solution is 28. Calculate the mole fraction (X) and molality (m) of the solute in the solution.

The correct option is B X = 0.2, m = 1259 Mol. of solute in 100 g solution =2828=1 Mol. of water in 100 g solution =100−2818=4 Mol. fraction of solute =11+4=0.2; Molality =1×100072=1259

The correct option is B
X = 0.2, m = $\frac{125}{9}$

Mol. of solute in 100 g solution $= \frac{28}{28} = 1$
Mol. of water in 100 g solution $= \frac{100 - 28}{18} = 4$
Mol. fraction of solute $= \frac{1}{1 + 4} = 0.2;$
Molality $= \frac{1 \times 1000}{72} = \frac{125}{9}$