Question

45g of ethylene glycol is mixed with 600g of water. What is the freezing point of the solution? ${\mathrm{k}}_{\mathrm{f}}=1.86\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}$

Answer 1

45 g of ethylene glycol is mixed with 600 g of water.
The molar mass of ethylene glycol =62 g/mol
Moles of ethylene glycol=$\frac{45}{62}$=0.726 moles.
600g of water =0.600kg of water.
Molality (m): The number of moles of solute dissolved in 1000g (1 kg) of a solvent is known as the molality of the solution.
$\mathrm{Molality}\left(\mathrm{m}\right)=\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}}{\mathrm{Mass}\mathrm{of}\mathrm{solvent}(\mathrm{in}\mathrm{Gram})}\times 1000$

$\mathrm{Molality}\left(\mathrm{m}\right)=\frac{0.726}{600}\times 1000=1.21\mathrm{m}$

The depression in the freezing point ΔT =${\mathrm{k}}_{\mathrm{f}}\times \mathrm{m}$
$∆\mathrm{T}={\mathrm{k}}_{\mathrm{f}}\times \mathrm{m}=1.86\times 1.21=2.25\mathrm{K}$
​The freezing point of pure water =273.15 K.
The freezing point of solution =273.15+2.25=270.90 K

45g of ethylene glycol is mixed with 600g of water. The freezing point of the solution is 270.90K.