Question

11. At a certain hill station, water
boils at 96 degree celsius. The amount
of NaCl that should be added to 1 litre
of water so that it boils at 100 degree
celsius will be (kb for water is 0.52k/m)
450 g
225 g
125 g
250 g​

Answer 1

Answer:

Sorry I didn't know the answer of this question

Answer 2

Given info : At a certain hill station, water

boils at 96° C.

To find : The amount of NaCl that should be added to 1 litre of water so that it boils at 100°C will be ...

solution : initial boiling point, T_i = 96°C

final boiling point, T_f = 100°C

so, change in boiling point, ∆T_b = T_f - T_i = 100°C - 96°C = 4°C

mass of 1 litre of water = 1 litre × 1 g/ml [ density of water is 1 g/ml ]

= 1000 ml × 1 g/ml = 1000g

let no of moles of NaCl is x

so, molality of NaCl and water solution, m = no of moles of NaCl/mass of solvent in Kg

= x/(1000/1000 kg) = x molal

using formula, ∆T_b = K_b × m

given, K_b for water = 0.52 k kg/molal

m = x molal

∆T_b = 4°C

∴ 4 = 0.52 × x

⇒4/0.52 = x

⇒x = 7.7

so the no of moles of NaCl = 7.7 moles

now mass of NaCl = moles × molar mass of NaCl

= 7.7 moles × 58.5 g/mol [ ∵ molar mass of NaCl is 58.5 g/mol ]

= 450.45 g ≈ 450 g (approx)

Therefore 450 g NaCl should be added to 1 litre of water so that it boils at 100°C.