Let, the molar mass of the solute be M g mol ^{- 1}

Now, the no. of moles of solvent (water),n₁ = 90g / 18g mol^-1

And, the no. of moles of solute,n₂ = 30g / M mol^-1 = 30 / M mol

p₁ = 2.8 kPa

Applying the relation:

(p₁⁰ - p₁) / p₁⁰    =  n₂ / (n₁  + n₂)

⇒ (p₁⁰ - 2.8) / p₁⁰    =  (30/M)  / {5  + (30/M)}

⇒ 1 - (2.8/p₁⁰)  = (30/M) / {(5M+30)/M}

⇒ 1 - (2.8/p₁⁰) = 30 / (5M + 30)

⇒ 2.8/p₁⁰  = 1 -  30 / (5M + 30)

⇒ 2.8/p₁⁰   =  (5M + 30 - 30) / (5M + 30)

⇒ 2.8/p₁⁰   = 5M / (5M+30)

⇒ p₁⁰ / 2.8  =  (5M+30) / 5M  ----------------(1)

After the addition of 18 g of water:

n₁ = (90+18g) / 18  = 6 mol

and the new vapour pressure is p₁ = 2.9 kPa   (Given)

Again, applying the relation:

(p₁⁰ - p₁) / p₁⁰    =  n₂ / (n₁  + n₂)

⇒ (p₁⁰ - 2.9) / p₁⁰    =  (30/M)  / {6  + (30/M)}

⇒ 1 - (2.9/p₁⁰)  = (30/M) / {(6M+30)/M}

⇒ 1 - (2.9/p₁⁰) = 30 / (6M + 30)

⇒ 2.9/p₁⁰  = 1 -  30 / (6M + 30)

⇒ 2.9/p₁⁰   =  (6M + 30 - 30) / (6M + 30)

⇒ 2.9/p₁⁰   = 6M / (6M+30)

⇒ p₁⁰ / 2.9  =  (6M+30) / 6M  ----------------(2)

Dividing equation (1) by (2),we get:

2.9 / 2.8 =   {(5M+30) / 5M} / {(6M+30) / 6M}

⇒ 2.9 x (6M+30 / 6)  =  (5M+30 / 5) x 2.8

⇒ 2.9 x (6M +30) x 5 = (5M+30) x 2.8 x 6

⇒ 87M + 435  =  84M + 504

⇒ 3M = 69

⇒ M = 23u

Therefore, the molar mass of the solute is 23 g mol - 1.

(ii) Putting the value of 'M' in equation (i), we get:

⇒ p₁⁰ / 2.8  =  (5M+30) / 5M 

⇒ p₁⁰ / 2.8  =  (5x23+30) / 5x23

⇒  p₁⁰  =  (145 x 2.8) / 115

⇒  p₁⁰  =  3.53

Hence, the vapour pressure of water at 298 K is 3.53 kPa.