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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.
उत्तर
(i) Let, the molar mass of the solute be M g mol−1.
Now, the no. of moles of solvent (water), n1 = `(90 "g")/(18 "g mol"^(-1))` = 5 mol
And, the no. of moles of solute, n2 = `(30 "g")/("M mol"^(-1)) = 30/"M" "mol"`
p1 = 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")/((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
`"n"_1 = (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")/((6"M" + 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 = 23u
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 = (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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