1.

A solution of sucrose (molar mass =342 g mol^(-1)) has been prepared by dissolving 68.5 g of sucrose in 100 g of water. The freezing point of the solution obtained will be (K_(f) "for water =1.86 K kg " mol^(-1)):

Answer»

`-0.372^(@)C`
`-0.52^(@)C`
`+0.372^(@)C`
`-0.57^(@)C`

Solution :MOLALITY (m) of solution
`=("Mass of solute 1 KG of solvent")/"MOLAR mass"`
`=((68.5g)//(1kg))/((342gmol^(-1)))=0.2 MOL//kg`
`DeltaT_(f)=K_(f)xxm`
` =(1.86 K kg mol^(-1))xx(0.2 mol//kg^(-1))`
f.p. of solution =`(0-0.372)=-372^(@)C`.


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