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A solution containing 30 g of non-volati...

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is added to this solution. The new vapour pressure becomes 2.9 kPa at 298 K. Calculate
(i) The molecular mass of solute and
(ii) vapour pressure of water at 298K

Text Solution

Verified by Experts

Let the vapour pressure of water at 298 K = `P^(0)`
the molecular Mass of solute = m
the molecular Mass of water = 18
mass of water = 90 gm
Mass of solute = 30 gm
Mole fraction of the non-volatile solute
= `(30//m)/((90)/(18) + (30)/(m)) = (30)/(30 + 5m) = 1 `
The vapour pressure of solute = 2.8 kPa at 298.
Now, `" " = (P^(0) - 2.8)/(P^(0)) = (30)/(30 + 5m)` ... (1)
Now, 18 gm of water is added to the solution
Now , mass of water = 108
Mole fraction of the solute
`= (30//m)/((180)/(18) + (30)/(m)) = (30)/(30 + 6 m )`
New relative lowering `(P^(0) - 2.9)/(p^(0))`
`= (30)/(30 + 6m) " " ` ... (2)
Soving equation (1) and (2) , we get
m = 34 g
Molecular mass of non-volatile solute = 34 g
Vapour pressure of water at 298 K = 3.4 kPa
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