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प्रश्न
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:
- molar mass of the solute.
- vapour pressure of water at 298 K.
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उत्तर
(i) Let the molar mass of the solute be M g mol−1.
Now, the number of moles of solvent (water), (n1) = `(90 g)/(18 g "mol"^-1)`
= 5 mol
And the number of moles of solute (n2) = `(30 g)/(M "mol"^-1) = 30/(M "mol")`
p1 = 2.8 kPa
By using Raoult’s law:
`(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)/((5 M + 30)/M)`
⇒ `1 - 2.8/p_1^0 = 30/(5 M + 30)`
⇒ `2.8/p_1^0 = 1 - 30/(5 M + 30)`
⇒ `2.8/p_1^0 = (5 M + 30 - 30)/(5 M + 30)`
⇒ `p_1^0/2.8 = (5 M + 30)/(5 M)` ...(i)
After the addition of 18 g of water
n1 = `(90 + 18 g)/18` = 6 mol
p1 = 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/(6 M + 30)`
⇒ `2.9/p_1^0 = 1 - 30/(6 M + 30)`
⇒ `2.9/p_1^0 = (6 M + 30 - 30)/(6 M + 30)`
⇒ `2.9/p_1^0 = (6 M)/(6 M + 30)`
⇒ `p_1^0/2.9 = (6 M + 30)/(6 M)` .......(ii)
Dividing equation (i) by (ii), we have:
`2.9/2.8 = ((5 M + 30)/(5 M))/((6 M + 30)/(6 M))`
⇒ `2.9/2.8 xx (6 M + 30)/6 = (5 M + 30)/5`
⇒ 2.9 × 5 × (6m + 30) = 2.8 × 5 × (5M + 30)
⇒ 87M + 435 = 84M + 504
⇒ 3M = 69
⇒ M = 23 u
Therefore, the molar mass of the solute is 23 g mol−1.
(ii) Putting the value of ‘M’ in equation (i), we have:
`p_1^0/2.8 = (5 xx 23 + 30)/(5 xx 23)`
`=> 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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