Question

Define molality $\left(m\right)$. Calculate molality $\left(m\right)$ of $10gm$ of Glucose $\left(C_6{H}_{12}{O}_6\right)$ in $90gm$ of water.

Answer 1

Molality $m=\frac{\text{number moles of solute}}{\text{mass of solvent in kg}}$
The molar mass of glucose is $6\left(12\right)+12\left(1\right)+6\left(16\right)=180$ g/mol. The mass of glucose (10 gm) is divided with molar mass (180 g/mol) to obtain the number of moles of glucose.
Number of moles n $=\frac{\text{10 g}}{\text{180 g/mol}}=\text{0.0555 mol}$
The mass of water $=\text{90 g}=\frac{\text{90 g}}{\text{1000 g/kg}}=\text{0.090 kg}$
Molality $\text{m}=\frac{\text{0.0555 mol}}{\text{0.090 kg}}=\text{0.62 m}$