Home
Class 12
CHEMISTRY
The weight of a non-volatile solute (mol...

The weight of a non-volatile solute (molecular weight = 40), which should be dissolved in 114g octane to reduce its vapour pressure to 80% is ___________ .

Text Solution

Verified by Experts

`(P^(@)-P_(S))/(P_(S)) = (n)/(N) = (wM)/(m xx W)`, Ocatne is `C_(8)H_(18)`.
(Given `m = 40, W = 114g, M_("Ocatne") = 114`)
`(100-80)/(80) = (w xx 114)/(40 xx 114)`
`w = 10 g`
Note: By `(P^(@)-P_(S))/(P_(S)) = (n)/(N)`
(Only for dilute solution, the answer comes `8 g`.)
Promotional Banner

Topper's Solved these Questions

  • DILUTE SOLUTION AND COLLIGATIVE PROPERTIES

    P BAHADUR|Exercise Exercise 3B Objectice Problems|1 Videos

  • DILUTE SOLUTION AND COLLIGATIVE PROPERTIES

    P BAHADUR|Exercise Exercise 9 Advanced Numerical|12 Videos

  • CHEMICAL KINETICS

    P BAHADUR|Exercise Exercise 9|13 Videos

  • ELECTROCHEMISTRY

    P BAHADUR|Exercise Exercise (9) ADVANCED NUMERICAL PROBLEMS|36 Videos

Similar Questions

Explore conceptually related problems

Calculate the weight of non - volatile solute having molecular weight 40, which should be dissolvd in 57 gm octane to reduce its vapour pressure to 80% :

The mass of a non-volatile solute (Molar mass 40) which should be dissolved in 114g octance to reduce its vapour pressure to 80% is 2xg . Find the value of x

The mass of a non-volatile solute of molar mass 40g" "mol^(-1) that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is

Mass of non-volatile solute "(X)" which should be, dissolved in "58g" of acetone to reduce its vapour pressure to "90%" will be (Atomic mass of "X=4" )