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​THE VAPOUR PRESSURE OF ACETONE AT 20 C IS 185 TORR. WHEN 1.2 G OF A NON – VOLATILE SUBSTANCE WAS DISSOLVED IN 100G OF ACETONE AT 20 C ITS VAPOUR PRESSURE WAS 183 TORR . THE MOLAR MASS OF THE SUBSTANCE IS

  1. ​THE VAPOUR PRESSURE OF ACETONE AT 20 C IS 185 TORR. WHEN 1.2 G OF A NON – VOLATILE SUBSTANCE WAS DISSOLVED IN 100G OF ACETONE AT 20 C ITS VAPOUR PRESSURE WAS 183 TORR . THE MOLAR MASS OF THE SUBSTANCE IS

Grade:11

2 Answers

BIKRAM
54 Points
6 years ago
This question is related to lowering of vapour pressure. Roult’s law P = X1×P0; X1=(183/185)=0.9891 ;P0=185 torr ,P=183 torr
: amount of acetone= 100/58 = 1.724 mol;amount of substance =1.2/M;
X1=1.724/(1.724 +1.2/M)= 0.9891;
1.724 +1.2/M =1.7429 ; M = 1.2/(1.7429-1.724) = 63.5
ankit singh
askIITians Faculty 614 Points
3 years ago
The molar mass M2​ of non-volatile substance can be calculated from the following formula.

p0p0−p​=W1​×M2​W2​×M1​​

Here, p0 vapour pressure of acetone and p is the vapour pressure of the solution.
M1​ and W1​ are the molar mass and mass of acetone.
M2​ and W2​ are the molar mass and mass of a non-volatile substance.

Substitute values in the above expression.

185185−183​=100×M2​1.2×58​

Hence, M2​=64.38g/mol.
Thus, the molar mass of the non-volatile substance is 64 g/mol (approximately).
Hence, the correct option is B

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