Calculate the number of moles of nonvolatile solute dissolved in 0.5 kg solvent if molal elevation constant for solvent is $2 \mathrm{~kg} \mathrm{~K} \mathrm{~mol}^{-1}$ $\left[\Delta \mathrm{T}_{\mathrm{b}}=0.8 \mathrm{~K}\right]$

The number of moles is determined using the boiling point elevation formula $\Delta T_b = K_b m$, with $\Delta T_b = 0.8\text{ K}$ and $K_b = 2\text{ kg K mol}^{-1}$.

Molality $m$ relates to moles $n$ and solvent mass $0.5\text{ kg}$ as $m = n / 0.5$.

Substituting into the elevation formula:
$0.8 = 2 \cdot \frac{n}{0.5}$

Solving for $n$:
$n = \frac{0.8 \times 0.5}{2} = \frac{0.4}{2} = 0.2\text{ mol}$

The number of moles of nonvolatile solute is $\boxed{0.2}$.