Answer to Question #319142 in General Chemistry for JBA

Question

The molal freezing point of water is 1.86°C/m (1.86K/m), which can be expressed as 1.86°C kg H₂O/mal solute (1.86K kg H₂O/mol solute) after some algebraic rearrangement. This means that a solution of 1 mol cane sugar (342g, over-Ib) dissolved in 1kg of water should freeze at -1.86°C.

Describe the freezing point drop suggestes in number 1.

Answer 1

$\Delta$ T_f = K_f * m(molality);

m(molality) = n(sugar)/m(water, kg).

n(sugar) = m(sugar)/M(sugar);

m(sugar) = 342 g;

M(sugar) = 342 g/mol;

K_f = 1.86 K/m;

T₀(water, H₂O) = 273 K = 0 °C

Therefore:

$\Delta$ T_f = (K_f * m(sugar))/(M(sugar)) * m(water, kg)) = (1.86 * 342)/(342 * 1) = 1.86 K = 1.86 °C;

$\Delta$ T_f = T₀ - T_f;

T_f = T₀ - $\Delta$ T_f = 0 - 1.86 = -1.86 °C.

Answer: -1.86 °C.