Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Vapour pressure of heptane p10 = 105.2 kPa
Vapour pressure of octane p20= 46.8 kPa
As we know that, Molar mass of heptane (C7H16) = 7 × 12 + 16 × 1 = 100 g mol - 1
∴ Number of moles of heptane = 26/100 mol = 0.26 mol
Molar mass of octane (C8H18) = 8 × 12 + 18 × 1 = 114 g mol - 1
∴ Number of moles of octane = 35/114 mol = 0.31 mol
Mole fraction of heptane, x1 = 0.26 / 0.26 + 0.31
= 0.456
And, mole fraction of octane, x2 = 1 - 0.456 = 0.544
Now, partial pressure of heptane, p1 = x2 p20
= 0.456 × 105.2
= 47.97 kPa
Partial pressure of octane,p2 = x2 p20
= 0.544 × 46.8 = 25.46 kPa
Hence, vapour pressure of solution, ptotal = p1 + p2
= 47.97 + 25.46
= 73.43 kPa
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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 16: Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid componen....
Comments
35/114=0.30 .....but why 0.31?
It was a suitable answer
There is a mistake for finding partial pressure. It should have been written Pa = xap°a
The yellow background soothes me.
dhanywad
God bless ur soul
Wow thanks
p1 = x2 p20??
Thanks for this help ....best &simple solution
Thanks for this help Best site!!!!!!