Question

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

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## 45 g of ethylene glycol is mixed with 600 g of water.The molar mass of ethylene glycol =62 g/molMoles 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}\left(\mathrm{in}\mathrm{Gram}\right)}Ã—1000$$\mathrm{Molality}\left(\mathrm{m}\right)=\frac{0.726}{600}Ã—1000=1.21\mathrm{m}$The depression in the freezing point Î”T =${\mathrm{k}}_{\mathrm{f}}Ã—\mathrm{m}$$âˆ†\mathrm{T}={\mathrm{k}}_{\mathrm{f}}Ã—\mathrm{m}=1.86Ã—1.21=2.25\mathrm{K}$â€‹The freezing point of pure water =273.15 K.The freezing point of solution =273.15+2.25=270.90 K45g of ethylene glycol is mixed with 600g of water. The freezing point of the solution is 270.90K.

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