Question

a sugar syrup of mass 214.2 containing 34.2 gram of sucrose calculate molality and mole fraction...??

Answer 1

214.2/34.2=9.10is your answer

Answer 2

Answer:

The molality of the solution is 0.555 mol/kg.

Mole fraction of water is 0.99.

Mole fraction of sucrose 0.0098.

Explanation:

Mass of sugar syrup solution = 214.2 g

Mass of sucrose = 34.2 g

Mass of solution = mass of solute + mass of solvent

214.2 g = 34.2 g + mass of solvent

Mass of solvent = 180 g = Mass of water

Moles of sucrose =$n_1=\frac{34.2 g}{342.3 g/mol}=0.0999 mol$

Moles of water =$n_2=\frac{180 g}{18 g/mol}=10 mol$

Mole fraction of sucrose =$\chi_1=\frac{n_1}{n_1+n+2}=\frac{0.0999 mol}{0.0999 mol+10 mol}=0.0098$

Mole fraction of water=$\chi_2=\frac{n_2}{n_1+n+2}=\frac{10 mol}{0.0999 mol+10 mol}=0.99$

$Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}$

$m=\frac{0.0999 mol}{0.180 kg}=0.555 mol/kg$

The molality of the solution is 0.555 mol/kg.