Class 12 Chemistry - Chapter Solutions NCERT Solutions | 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 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.
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.
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4 Comment(s) on this Question
Thank you so much..it really helped me a lot
Why are you adding 1 in p°-p/p° ??
Nice
Really good
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