Question

13/14. The vapor pressure of water at 20 °C is 17.5 mmHg. What is the vapor pressure of water over a solution prepared from 2.00 x 102 g of sucrose (C12H22011) and 3.50 x 102 g water at 20°C? (a) 0.51 mmHg (b) 16,0 mmHg ) 17.0 mmHg (d) 18.0 mmHg (e) 19.4 mmHg 15/16. Determine the freezing point of a solution which contains 0.31 mol of sucrose in 175 g of water (Kr= 1.86 °C/(mol-kg-'). (a) -3.3 °C (b)-1.1 °C 0.0°C (d) 1.1 °C (e) 3.3 °C

Answer 1

13/14)

Mass of sucrose 200gm

Molar mass of sucrose 342 gm /mol

Moles of sucrose =mass taken/molar mass=200/342=0.58

Mass of water taken 350gm

Molar mass of water 18gm

Moles of water =mass taken/molar mass= 350gm/18gm=19.44

Mole fraction of water = moles of water/(moles of water +moles of sucrose) =19.44/(19.44+0.58)=19.44/20.02=0.971

According to the Raoults law

Vapour pressure of solution =mole fraction of water&vapour pressure of pure water =0.971*17.5 mm of Hg=16.99 mm of Hg (nearly 17 mm of Hg) option 3 is correct

15/16

Moles of sucrose 0.31

Mass of water 175g =0.175 kg (1kg =1000g)

Molality =moles of sucrose/ mass of water= 0.31moles/0.175kg=1.77 m

Depression in freezing point is given by

$\Delta T{_{f}=K^{_{f}}*molality$

So

$\Delta T{_{f}=1.86 ^{0C}*kg^{-1}*mol*1.77mol/kg$=3.3⁰C

So freezing point of solution =freezing point of pure water -depression in freezing point=(0-3.3)⁰C=-3.3⁰C

option a is correct