Question

What mass of ethylene glycol must be added to 5.50kg of water to lower the freezing point of water from 0°C to -10°C?

Answer 1

Answer: 1833.3g

Explanation: Formula used for lowering in freezing point is,

$\Delta T_f=k_f\times m$

$T_f$ = change in freezing point= $10^0C$

$k_f$ = freezing point constant = $1.86^0C/m$

m = molality

where molality is defined as the moles of solute dissolved per kg of the solvent.

$\Delta T_f=k_f\times \frac{\text {Given mass}}{\text{ Molar mass}\times \text{ Mass of solvent in kg}}$

$10^0C=1.86^0C/m\times \frac{\text {Given mass}}{62g/mol}\times 5.50kg$

${\text {Given mass}}=1833.3g$