Question

Homemade ice-cream is frozen by churning it in a bucket suspended in an ice-water-salt mixture. A typical mix calls for 1.80 kg of salt (NaCl) and 5.00 kg of ice. Compute the mole fraction of NaCl in this mixture after all the ice melts. Compute the freezing point of this mixture (in °C).

Answer 1

Ans :-

No. of moles of NaCl (nNaCl) = Given mass of NaCl / Gram molar mass of NaCl

= 1.80 x 103 g / 58.44g/mol

= 30.8 mol

No. of moles of Ice (nH2O) = Given mass of Ice / Gram molar mass of Ice

= 5.00 x 103 g / 18 g/mol

= 277.8 mol

Therefore,

Mole fraction of NaCl (XNaCl) = nNaCl/nNaCl + nH2O

= 30.8 mol / (30.8 + 277.8) mol

= 0.0998

Therefore, Mole fraction of Mole fraction of NaCl (XNaCl) = 0.0998

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From Depression in freezing point (ΔTf) :

ΔTf = Tf0-Tf = i.kf.m ......................(1)

Where, Tf0 = Freezing point of ice or water = 0 °C

Tf = Freezing point of ice or water = ?

i = Vant's hoff factor of NaCl = 2 as NaCl gives two ions in aqueous solution one is Na+ and one is Cl-.

m = Molality of NaCl = Number of moles of NaCl / Mass of NaCl in kg

= 30.8 mol / 5.00 kg

= 6.16 mol/kg or 6.16 m

and

kf = Molal freezing point constant of water = 1.86 °C/m

Put all these values in equation (1) :

Tf0-Tf = i.kf.m

0 °C - Tf = (2).(1.86 °C/m).(6.16 m)

- Tf = 22.9 °C

Tf = -22.9 °C

Therefore, Freezing point of mixture = -22.9 °C