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Academic session 2026-27, Chapter 1 in the current curriculum.
Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter Solutions. This page offers a step-by-step solution to the specific question from Exercise 2, Question 18:
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%.
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.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.
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate/mol L - 1 min - 1 |
| I | 0.1 | 0.1 |
2.0 × 10 - 2 |
| II | -- | 0.2 |
4.0 × 10 - 2 |
| III | 0.4 | 0.4 | -- |
| IV | -- | 0.2 |
2.0 × 10 - 2 |
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 18: Calculate the mass of a non-volatile solute (molar mass 40 g mol-1) which should be dissolved in 114....
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Comments
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= 80ÃâPo/100 = 0.8Po Ps =Po Ãâ 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=PoÃâ1/(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