Question

What is the boiling point, in °C,

of a 0.321 molal aqueous
solution of NaCl?
BP(water) = 100 °C
Kb (water) = 0.512 °C/m
[?] °C

Answer 1

the boiling point of solution at 3 decimal point is 100.329०C Ans.

Step-by-step explanation:

given data -

molality of Nacl = 0.321 m

molal boiling point elevation constant (Kb) =0.512०C/m

# formula of change of boiling point of sample =

∆ Tb =i × Kb × m

Kb = molal boiling point of elevation constant

m = molality

i = vont's hoff factor.

Nacl is strong electrolyte and its 100% dissociate so the value of i for Nacl is 2

put value in the formula

∆ Tb = 2 × 0.512 ०C/m × 0.321m

= 0.3287

= 0.329०C

∆Tb = T'b - Tb

T'b = boiling point of solution

Tb= boiling point of solvent( water)

0.329०C = T'b - 100०c ( boiling point of water = 100०C)

T'b = 0.329०C + 100०C

= 100.329०C