Calculate the mass of a non-volatile solute (molar mass 40 g mol-1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Let the vapour pressure of pure octane be p10.
Then, the vapour pressure of the octane after dissolving the non-volatile solute is 80/100 p10 = 0.8 p10.
Molar mass of solute, M2 = 40 g mol - 1
Mass of octane, w1 = 114 g
Molar mass of octane, (C8H18), M1 = 8 × 12 + 18 × 1 = 114 g mol - 1
Applying the relation,
(p10 - p1) / p10 = (w2 x M1 ) / (M2 x w1 )
⇒ (p10 - 0.8 p10) / p10 = (w2 x 114 ) / (40 x 114 )
⇒ 0.2 p10 / p10 = w2 / 40
⇒ 0.2 = w2 / 40
⇒ w2 = 8 g
Hence, the required mass of the solute is 8 g.
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10 Comment(s) on this Question
Give the complete & correct solution, immediately
This solution is quite easy but this is not correct. The correct answer for this question is 10.
This is quite easy but correct answer is 10
For the answer to be 10 , it should be equated with X of solute i.e. ((w1/M1)/(w1/M1)+(w2/M2))
Correct is 10
But answer is 10
Give the correct solution
Answer is right
The answer is wrong ....the correct answer should be done like given below if vapour pressure of pure liquid is = Po 80 % of pure liquid Ps= 80ÃPo/100 = 0.8Po Ps =Po à Xsolute mass of solute = x gram mass of solvent = 114g Molar mass of solute= 40 g/mol Molar mass of solvent (octane C8H18) = 114g/mol Number of moles of solute = x/40 = 0.025x Number of moles of solvent = 114/114= 1 moles Mole fraction of solvent = 1/(1+0.025x) 0.8Po=PoÃ1/(1+0.025x) Cross multiply we get (1+0.025x))0.8Po= Po Divide by 0.8 Po we get 1+0.025x = 1.25 Subtract 1 both side we get 0.025x = 0.25 Now divide by 0.025 we get x = 10g
this answer in wrong instead the formula used will be p-ps/ps
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