Question

The boiling point of water is 100
c. Calculate thr boiling point of an aqueous solution containing 0.6 g of urea in 100 kg of water

Answer 1

Explanation:

from elevation in boiling,

=kb.m

where \Delta T_bΔT

b

is elevation in boiling point, K_bK

b

is ebullioscopic constant and m is molality of solute.

mass of urea = 0.6 g

molar mass of urea (solute) = 60g/mol

so, number of mole of urea = 0.6/60 = 0.01

mass of solvent (water) = 100g

so, molality of urea = number of mole of urea/mass of solvent in kg

= 0.01 × 1000/100 = 0.1 molal

given, ebullioscopic constant, K_bK

b

= 0.52K/molal or 0.52°C/molal

then, ∆Tb = 0.52 × 0.1 = 0.052K

so, boiling point of solution, Tb = T + 0.052°C

= 100°C + 0.052 °C = 100.052°C

Answer 2

Answer:

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