Question

A mixture weighing 228g contain CaCl2 and NaCl. If this mixture is dissolved in 10kg of water and form ideal solution that boil at 100.364°C. The mol% of NaCl in mixture is [kb of water is 0.52K mol-1 kg]​

Answer 1

we know, boiling point of water , T = 100°C

here given, boiling point of mixture , T' = 100.364°C

so, elevation in boiling point of water , ∆Tb = 100.364°C - 100°C = 0.364°C or 0.364K [ change in temperature remains same in both degree or Kelvin ]

from formula,

∆Tb = Kb × m , where m is molality.

given, Kb = 0.52K. kg/mol

so, 0.364 = 0.52 × m

or, m = 0.364/0.52 = 0.7 molal

0.7 molal means 0.7 mole of solute in 1kg of solvent.

mass of solvent is 10kg .so, number of mole of solute = 10 × 0.7 = 7 mole

now, Let in NaCl and CaCl2 mixture x mole of NaCl and y mole of CaCl2 are mixed in water.

so, mass of mixture of NaCl and CaCl2 = 58.5x + 111y = 228g

we get, x = 2 and y = 1

means, amount of NaCl is twice the amount of CaCl2.

now, n(x + y) = 7

or, n(2 + 1) = 7 => n = 7/3

then, nx = 14/3 and ny = 7/3

so, mol % of NaCl = 14/3/(14/3 + 7/3) × 100 = 14/21 × 100 = 200/3 = 66.7%