Home
Class 12
CHEMISTRY
A 5% solution (by mass) of cane sugar in...

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

Text Solution

Verified by Experts

(i) 5% solution of cane sugar `(C_(12)H_(22)O_(11))` contains 5g of sugar in 95 g of water.
Amount of sugar in 1 kg of water `=(5)/(95)xx1000= 52.63 g`
Molar mass of sugar,
`C_(12)H_(22)O_(11)=12xx12 +22xx1+16xx11=342"g mol"^(-1)`
Number of moles of sugar in 1 kg of water,
`m=(52.63)/(342)=0.1539`
Depression in F.P.,
`DeltaT_(f)=K_(f) xx m=273.15-271=2.15K`
`:.2.15=K_(f)xx0.1539`
`K_(f)=(2.15)/(0.1539)=13.97"K kg mol"^(-1)`
(ii) 5% solution of glucose contains 5g of glucose `(C_(6)H_(12)O_(6))` in 95 g of water.
Molar mass of glucose
`(C_(6)H_(12)O_(6)) = 6 xx 12 + 12xx 1 +6xx 16 = 180" g mol"^(-1)`
Amount of glucose in 1 kg of water
`=(5)/(95)xx1000=52.63g=(52.63)/(180)` moles
`=0.292"mol kg"^(-1)`
Molality of solution, `m=0.292 " mol kg"^(-1)`
`DeltaT_(f)=K_(f) xx m=13.97 xx 0.292=4.08K`.
Freezing point of 5% solution of glucose `= 273.15-4.08=269.07K`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    ICSE|Exercise Follow up Problems |58 Videos

  • SOLUTIONS

    ICSE|Exercise EXERCISE (PART-I Objective Questions)|26 Videos

  • SOLID STATE

    ICSE|Exercise ISC EXAMINATION QUESTIONS PART-I (Numerical Problems)|6 Videos

  • SOME IMPORTANT ORGANIC NAME REACTIONS

    ICSE|Exercise Questions |278 Videos

Similar Questions

Explore conceptually related problems

(a) A 10% solution (by mass) of sucrose in water has freezing point of 269.15K. Calculate the freezing point of 10% glucose in water, if the freezing point of pure water is 273.15K. Given: (molar mass of sucrose=342 g mol^(-1) ) (Mola mass of glucose =180g mol^(-1) ) (b) Define the following terms: (i) Molality (m) (ii) Abnormal molar mass

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.

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

A 4% solution (w/w) of sucrose (M 342 g mol^(-1) ) in water has a freezing point of 271.15K Calculate the freezing point of 5% glucose (M= 180 g mol^(-1) ) in water. (Given: Freezing point of pure water 273.15 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 ?

The freezing point of solution is ……..than the boiling point of pure solvent.

A solution of urea in water has a boiling point of 100*18^@C . Calculate the freezing point of the solution.(K for water is 1*86 " K kg " mol^(-1) and K_b " for water is " 0*512 " K kg " mol^(-l) ).

Freezing point of solution is marked as

Solution with highest freezing point is