Calculate the mass of a non-volatile sol | Class 12 Chemistry Chapter Solutions, Solutions NCERT Solutions

Q18.

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 ) / (Mw1 )

(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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What is the correct answer to: 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 octan...

How do you solve 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%. step by step?

Step-by-step explanation:
• 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

What common mistakes should be avoided in this question?

Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.

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Student Discussion

Pankaj
Class · · , · May 24, 2019
Give the complete & correct solution, immediately
👍 👎 Reply
Tanya
Class · · , · May 04, 2019
This solution is quite easy but this is not correct. The correct answer for this question is 10.
👍 👎 Reply
Tanya
Class · · , · May 04, 2019
This is quite easy but correct answer is 10
👍 👎 Reply
Bhagya
Class · · , · Apr 25, 2019
For the answer to be 10 , it should be equated with X of solute i.e.
((w1/M1)/(w1/M1)+(w2/M2))
👍 👎 Reply
Gautam
Class · · , · Apr 22, 2019
Correct is 10
👍 👎 Reply
Neha
Class · · , · Mar 28, 2019
But answer is 10
👍 👎 Reply
Neha
Class · · , · Mar 28, 2019
Give the correct solution
👍 👎 Reply
Renuka
Class · · , · Mar 10, 2019
Answer is right
👍 👎 Reply
Malavika Manish
Class · · , · Mar 09, 2019
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= 80×Po/100 = 0.8Po
Ps =Po × 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=Po×1/(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
👍 👎 Reply
nikhil gupta
Class · · , · Jan 17, 2019
this answer in wrong instead the formula used will be p-ps/ps
👍 👎 Reply

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