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

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

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X = 0.8, m = 10

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X = 0.2, m = 10

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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 $= \frac{1 \times 1000}{72} = \frac{125}{9}$

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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 $= \frac{1 \times 1000}{72} = \frac{125}{9}$