Question

0.02 M of cane sugar C12H22O11 solution is prepared in 400 ml of the solution. How much weight of cane sugar is used?​

Answer 1

Explanation:

A 5% solution (by mass) of cane sugar in water has ... The depression in the freezing point of the solution is given ... Molality of sucrose solution =0.0950.0146=0. 154 m

Missing: 0.02 ‎400 ‎ml

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Answer 2

Answer:

The depression in the freezing point of the solution is given by

ΔT

f

= freezing point of water − freezing point of solution

ΔT

f

=273.15−271=2.15 K

Molar masses of glucose and sucrose are 180 g/mol and 342 g/mol respectively.

100 g of solution will contain 5 g of glucose or 5 g of sucrose.

Number of moles of glucose =

180

5

=0.028 moles

Number of moles of sucrose =

342

5

=0.0146 moles

Mass of solvent = total mass of solution - mass of solute =100−5=95 g or 0.095 kg

Molality of sucrose solution =

0.095

0.0146

=0.154 m

K

f

=

molality

ΔT

f

=

0.154

2.15

=13.97

For glucose solution, ΔT

f

=K

f

×m=13.97×0.29=4.08

Freezing point of 5% glucose solution in water = freezing point of water - ΔT

f

=273.15−4.08=269.07 K