In a solution of 90 g water, 30 g of non-volatile solute is present which has exactly a vapor pressure of 2.8kPa at 298 K. On further addition of 18 g water to the solution, the new vapour pressure of the solution becomes 2.9 kPa at 298 K. Determine (i) the molecular weight of the solute (ii) vapor pressure of water at 298 K

Options

A. `23"g mol"^(-1), 3.53`
B. `26"g mol"^(-1), 3.53`
C. `28"g mol"^(-1), 3.53`
D. `32"g mol"^(-1), 3.53`

Answer

A

Explanation

Step I: Calculation of molar mass of the solute
Case I: Number of mole of solute:
`(n_(B))=("Mass")/("Molar mass") =((30g))/((Mg mol^(-1))) =(30)/(M)` mol
Number of moles of water:
`(n_(A)) =("Mass")/("Molar mass") =((90g))/((18g mol^(-1)))=5` mol
Molar fraction of water:
`(X_(A))=(n_(A))/(n_(A)+n_(B)) =((5mol))/((5mol +(30)/(M)"mol")) =(M)/((6+M))`
`{X_(A)= "mole traction of water"}`
Vapour pressure of 1st solution `(p_(A))=2.8`kPa
According to Raoult.s law, `p_(A)""^(0)X_(A)`
`(2*8 kPa) =p_(A)^(o) x_(A) =p_(A)^(o) (M)/(6+M)" "(i)`
Case II: Number of moles of solute
`(n_(B)) =(30)/(M)` mol
The number of moles of water:
`(n_(A)) =("Mass")/("Molar mass") =((108g))/((18g mol^(-1)))=6` mol
Mole fraction of water `(X_(A))=(n_(A))/(n_(A)+n_(B))`
`=((6mol))/((6mol +(30)/(M)"mol"))`
`=(M)/((5+M))`
Vapour pressure of the solution `(p_(A))=2.9`kPa
According to Raoult.s law, `p_(A) =p_(A)""^(o)X_(A)`
`(2.9kPa)=p_(A)^(o) xx (M)/((5+M)) " " ...(ii)`
Dividing Eq. (i) by Eq. (ii)
`((2.8kPa))/(( 2.9kPa)) =((5+M))/((6+M))`
`0.9655 =((5+M))/((6+M))`
`(0.9655 xx 6)+ 0.9655M=5+M`
`0.0345M=0.793`
`M=(0.793)/(0.0345)=23g mol^(-1)`
Step II: Calculation of vapour pressure of water
According to Rauolt.s law, `p_(A)=p_(A)""^(o)X_(A)`
`2.8kPa = p_(A)^(o) xx (M)/(6+M) " " ....(iii)`
Placing the value of M in Eq. (i)
`2.8kPa =p_(A)^(o) (23)/(6+23)`
`p_(A)^(o) =(2.8 xx 29)/(23)`
`=(81.2)/(23) = 3.53kPa`