Question

The boiling point constant for water is 0.51 Kkg/mol. Calculate boiling point of water in a solution gf 0.010 mol/kg of sugar and a solution of 0.010 mol/kg of NaCl.

Answer 1

Sol :-

From elevation in boiling point (which is a colligative property) expression :

ΔT_b = T_b - T⁰_b = i.K_b.m .................(1)

Here,

T_b = Boiling point of water in a solution

T⁰_b = Boiling point of pure water = 100 ⁰C = 373.15 K

i = Vant's Hoff factor, (For NaCl, i = 2 because NaCl gives two ions in solution and for sugar (a non-electrolyte) i = 1.

m = Molality of solution

Calculation of boiling point of water in a solution gf 0.010 mol/kg of sugar :-

From equation (1) :

T_b - 373.15 K= 1 x 0.51 Kkg/mol x 0.010 mol/kg

T_b = 0.0051 K + 373.15 K

T_b = 373.1551 K

Hence,

Boiling point of water in a solution of 0.010 mol/kg of sugar = 373.1551 K

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Calculation of boiling point of water in a solution of 0.010 mol/kg of NaCl​​​​​​​:-

From equation (1) :

T_b - 373.15 K= 2 x 0.51 Kkg/mol x 0.010 mol/kg

T_b = 0.0102 K + 373.15 K

T_b = 373.1602 K

Hence,

Boiling point of water in a solution of 0.010 mol/kg of NaCl = 373.1602 K ​​​​​​​

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