Correct Answer - a.Mw2=22.71gmol-1 b.P∘=3.54kPa
n2=
30g
Mw2
,n1=
90g
18gmol-1
=5mol
Use the relation,
p∘-pS
pS
=
n2
n1
⇒
p∘-2.8
2.8
=
30/Mw2
5
⇒5×(
p∘-2.8
2.8
)=30/Mw2….(i)
After adding 18gH2O(i.e.,1molofH2O),
n1=6mol
∴
P∘-2.9
2.9
=
30/Mw2
6
⇒6×(
p∘-pS
2.9
)=30/Mw2.....(ii)
Equates Eqs.(i)and (ii)
5(
p∘-2.8
2.8
)=6(
p∘-2.9
2.9
)
Solve for p∘:
p^(@)=2.54kPa
Substitute the value of p^(@) either in Eq. (i) or in Eq. (ii).
5xx((3.54-2.8)/(2.8))=(30)/(Mw_(2))
Solve for Mw_(2),
:. Mw_(2)=22.71g mol^(-1)
