Question

A water sample contains 9.5% mgcl2 and 11.7% nacl (by weight). Assuming 80% ionisation of each salt. Boiling point of water will be approximately (kb = 0.52)

Answer 1

Answer:

377k

Explanation:

Assume 100 g soln

m(nacl)=11.7g

m(mgcl2)=9.5g

Mass of solvent (h2o)=100-11.7-9.5=78.8g

n(nacl)=11.7/58.5=0.2

n(mgcl2)=9.5/95=0.1

i-van't hoff factor

a-no of moles after dissociation

i of nacl

1+a=1.8

i of mgcl2

1+2a=2.6

∆T=Kb*1000*i*n2/(w1)

∆T=∆T(nacl)+∆T(mgcl2)

=0.52*1000*(1.8*0.2+2.6*0.1)/78.8=4.09

T(boiling)=T+∆T=373+4.09=377K

Answer 2

Assume 100 g soln

m(nacl)=11.7g

m(mgcl2)=9.5g

Mass of solvent (h2o)=100-11.7-9.5=78.8g

n(nacl)=11.7/58.5=0.2

n(mgcl2)=9.5/95=0.1

i-van't hoff factor

a-no of moles after dissociation

i of nacl

1+a=1.8

i of mgcl2

1+2a=2.6

∆T=Kb*1000*i*n2/(w1)

∆T=∆T(nacl)+∆T(mgcl2)

=0.52*1000*(1.8*0.2+2.6*0.1)/78.8=4.09

T(boiling)=T+∆T=373+4.09=377K