Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# 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 then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. - Chemistry

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 then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

(1) molar mass of the solute

(2) vapour pressure of water at 298 K.

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#### Solution

Let, the molar mass of the solute be M g mol−1

Now, the no. of moles of solvent (water),  n_1 = (90g)/18 g mol^(-1) = 5 mol

And, the no. of moles of solute,  n_2 = (30g)/(M mol^(-1)) = 30/M mol

p_1 = 2.8 kPa

Applying the relation:

(p_1^0-p_1)/p_1^0 = n_2/(n_1+n_2)

=> (p_1^0 - 2.8)/p_1^0 = (30/M)/(5 + 30/M)

=>1-2.8/p_1^0= (30/M)/((5M+30)/M)

=> 1-2.8/p_1^0 = 30/(5M+30)

=> 2.8/p_1^0=1-30/(5M+30)

=>2.8/p_1^0 = (5M+30-30)/(5M+30)

=>p_1^0/2.8 = (5M+30)/5M ......(i)

After the addition of 18 g of water

n_1 = (90+18g)/18 = 6 mol

p_1 = 2.9 kPa

Again, applying the relation:

(p_1^0-p_1)/p_1^0=n_2/(n_1+n_2)

=>(p_1^0-2.9)/p_1^0 = (30/M)/((6+30)/M)

=>1-2.9/p_1^0  = (30/M)/((6M + 30)/M)

=>1-2.9/p_1^0 = 30/(6M+30)

=>2.9/p_1^0 = 1 - 30/(6M+ 30)

=>2.9/p_1^0= (6M+30-30)/(6M+30)

=>2.9/p_1^0 = (6M)/(6M+30)

=>p_1^0/2.9 = (6M+30)/(6M) .......(ii)

Dividing equation (i) by (ii), we have:

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

=>2.9/2.8 xx (6M+30)/6 = (5M+30)/5

=> 2.9 x 5 x (6m + 30) = 2.8 x  5 x (5M + 30)

=> 87M + 435 = 84M + 504

=> 3M = 69

=> M = 23u

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

(2)Putting the value of ‘M’ in equation (i), we have:

p_1^0/2.8 = (5xx23+30)/(5xx23)

=> p_1^0/2.8   = 145/115

=> P_1^0 = 3.53`

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

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#### APPEARS IN

NCERT Class 12 Chemistry Textbook
Chapter 2 Solutions
Q 19 | Page 60
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