Question

1.45 g of sucrose (C$_{12}$H$_{22}$O$_{11}$) is dissolved in 30.0 mL of water. Molality (rounded off to 3 decimals) of the resulting solution is _______ m.

Hint:

Molality depends on mass of solvent (in kg), not on solution volume.

Answer 1

Correct Answer: 0.14

Step 1: Given data.
Mass of solute = 1.45 g
Molecular weight of sucrose = 342 g/mol
Mass of solvent = 30.0 mL of water = 30.0 g = 0.030 kg
Step 2: Calculate moles of solute.
\[ \text{Moles of sucrose} = \frac{1.45}{342} = 0.00424~\text{mol} \] Step 3: Calculate molality.
\[ m = \frac{0.00424}{0.030} = 0.141~\text{mol/kg} \] Step 4: Conclusion.
Molality of the solution = 0.142 m (rounded to 3 decimals).