Use freezing point depression formula

$\Delta T_f = i K_f m$

where:

• $\Delta T_f = 0.2\,^\circ\mathrm{C}$
• $K_f = 2\,\mathrm{K\,kg\,mol^{-1}}$
• for $\mathrm{NaCl}$, assuming complete dissociation: $i = 2$

Calculate molality required

$m = \frac{\Delta T_f}{iK_f} = \frac{0.2}{2\times 2} = 0.05\,\mathrm{mol\,kg^{-1}}$

Mass of solvent

Given water = $600\,\mathrm{g} = 0.6\,\mathrm{kg}$

Moles of NaCl needed

$n = m \times \text{kg of solvent} = 0.05 \times 0.6 = 0.03\,\mathrm{mol}$

Convert moles to mass

Molar mass of NaCl: $M = 23 + 35.5 = 58.5\,\mathrm{g\,mol^{-1}}$

Therefore, $w = nM = 0.03 \times 58.5 = 1.755\,\mathrm{g}$

Final answer

Required amount of NaCl: $\boxed{1.76\,\mathrm{g}}$

Comparison with stored correct answer

Stored correct answer: $1.74$ to $1.76$

Our answer $1.755\,\mathrm{g}$ lies in this range, so it agrees.