Question

Calculate the molality of a sugar solution that contains 3.94 g sucrose (C12H22O11) dissolved in 285 g water.

Answer 1

Molality= moles of solute /kg of solvent

moles of solute= mass/molar mass
moles of solute= 3.94 g / (342 g/mol)
moles of solute= 0.01 mol

molality= 0.01 mol / 0.285 kg
molality= 0.04 m

Answer 2

Answer : The molality of sugar solution is, 0.04038 mole/Kg

Solution : Given,

Mass of solute (sucrose) = 3.94 g

Mass of solvent (water) = 285 g

Molar mass of sucrose = 342.3 g/mole

Molality : It is defined as the number of moles of solute present in one kilogram of solvent.

Formula used :

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent in g}}=\frac{3.94g\times 1000}{342.3g/mole\times 285g}=0.04038mole/Kg$

Therefore, the molality of sugar solution is, 0.04038 mole/Kg