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A 5% solution (by mass) of cane sugar in...

A `5%` solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose (by mass) in water. The freezing point of pure water is 273.15 K.

Text Solution

Verified by Experts

`5%` solution by mass means 5 g of solute is present in 100 g of solution
`therefore `Mass of solvent (Water) = 95 g
`"Molality of sugar solution "=(5)/(342)xx(1000)/(95)=0.154" "("mol. Mass of sugar "C_(12)H_(22)O_(11)=342)`
`DeltaT_(f)" for sugar solution "=273.15-271=2.15^(@), DeltaT_(f)=K_(f)xxm therefore K_(f)=(2.15)/(0.154)`
`"Molality of glucose solution "=(5)/(180)xx(1000)/(95)=0.292" "("mol. mass of glucose "C_(6)H_(12)O_(6)=180)`
`therefore" "DeltaT_(f)" (Glucose)" =K_(f)xxm=(2.15)/(0.154)xx0.292=4.08`
`therefore" Freezing point of glucose solution "=273.15-4.08="269.07 K."`
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A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose in water if freezing point of pure water is 273.15 K.

A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose solution in water, if freezing point of pure water is 273.15 K. Given : (Molar mass of sucrose = "342 g mol"^(-1) , molar mass of glucose ="180 g mol"^(-1) ).

Knowledge Check

  • A 5% solution (by mass) of cane sugar in water has freezing point of 271 K and freezing point of pure water is 273.15 K. The freezing point of a 5% solution (by mass) of glucose in water is :

    A
    271 K
    B
    273.15
    C
    269.07 K
    D
    277.23 K.

  • A 5% solution (w/W) of cane sugar (molar mass = 342 g mol^(-1) ) has freezing point 271 K. What will be the freezing point of 5% glucose (molar mass = 18 g mol^(-1) ) in water if freezing point of pure water is 273.15 K ?

    A
    273.07 K
    B
    269.07 K
    C
    273.15 K
    D
    260.09 K

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