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A solution containing 30g of non-volatile solute exactly in 90g of water has a vapour pressure of 2.8 kpa at 298 k. further, 18g of water is then added to the solution & the new vapour pressure becomes 2.9 kpa at 298 k. Calculate.

a. Molar mass of the solute b. Vapour pressure of water at 298k

) The relative lowering of vapour pressure is given by the following expression

(p^{0}_{solvent} - p_{solution})/ p^{0}_{solvent} = n_{2}/(n_{1} + n_{2})

Where p^{0}_{solvent} is the vapour prseure of the pure solvent, p_{solution} is the vapour pressure of solution containing dissolved solute, n_{1} is the number of moles of solvent and n_{2} is the number of moles of solute.

For dilute solutions, n_{2} << n_{1}, therefore the above expression reduces to

(p^{0}_{solvent} - p_{solution})/ p^{0}_{solvent} = n_{2}/n_{1}

= (w_{2} X M_{1}) / (M_{2} X w_{1}) (i)

Where w_{1} and w_{2} are the masses and M_{1} and M_{2} are the molar masses of solvent and solute respectively.

We are given that

w_{2} = 30g

w_{1} = 90g

p_{solution} = 2.8kpa

p^{0}_{solvent } = ? and M_{2} = ?

Substituting these values in relation (i), we get

(p^{0}_{solvent} -2.8)/ p^{0}_{solvent} = (30 X 18)/ (M_{2} X 90)

(p^{0}_{solvent} -2.8)/ p^{0}_{solvent} = 6/ M_{2} (1)

Similarly for second case we have the following values

w_{2} = 30g

w_{1} = (90 + 18)g = 108g

p_{solution} = 2.9 kpa

Therefore we get

(p^{0}_{solvent} -2.9)/ p^{0}_{solvent} = (30 X 18)/ (M_{2} X 108)

= 5/M_{2} (2)

Dividing (1) by (2), We get

(p^{0}_{solvent} -2.8)/ (p^{0}_{solvent} -2.9) = 6/5

Therefore

p^{0}_{solvent} = 3.4kpa

That is vapour pressure of water at 298K is 3.4 kpa.

Substituting the value of p^{0}_{solvent} in (1), we get

(3.4- 2.8)/3.4 = 6/M_{2}

or 0.6/3.4 = 6/ M_{2}

Therefore M_{2} = 34g

Therefore mass of solute is 34g and vapour pressure of water at 298K is 3.4 kpa.

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