An ideal solution was prepared by dissolving some amount of can sugar (non-volatile) in 0.9 moles of water.The solution was then cooled just below its

Question

An ideal solution was prepared by dissolving some amount of can sugar (non-volatile) in 0.9 moles of water.The solution was then cooled just below its

An ideal solution was prepared by dissolving some amount of can sugar (non-volatile) in 0.9 moles of water.The solution was then cooled just below its freezing temperature (271 K) where some ice get separated out.The remaining aqueous solution registered a vapour pressure of 700 torr at 373K.Calculate the mass of ice separated out, if the molar heat of fusion of water is 6 kJ.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

`K_f=(RT_f^2)/(DeltaH_("fusion"))M`(M=mol. wt. `T_f` =normalfreezing pt)
`K_f=(8.314xx(273)^2xx18)/(1000xx6xx10^3)`
`:. K_f ~~1.86K.kg mol^(-1)`
Now `DeltaT_f=K_fxxm`(m=molality)
`:.m=(DeltaT_f)/K_f=((273-271))/1.86=1.07` moles/kg
But `m=("moles of solutes")/("weight of solvent (in Kg)")`
n=moles of solute =`1.075xx((0.9xx18)/1000)=1.74xx10^(-2)`
Also `X_("solute")=(760-700)/760=0.079` (where X=mole fraction)
`:. "Total moles"=(1.74xx10^(-2))/0.079=0.22`
Moles of solvent `(H_2O)=0.2026`
`:.` Mass of ice separated out =(0.9-0.2026)x18=12 gm