Home
Class 12
CHEMISTRY
Twenty grams of a solute are added to 10...

Twenty grams of a solute are added to `100g` of water at `25^(@)C`. The vapour pressure of pure water is `23.76 mmHg`, the vapour pressure of the solution is `22.41` Torr.
(a) Calculate the molar mass of the solute.
(b) What mass of this solute is required in `100g` of water of reduce the vapour pressure ot one-half the value for pure water?

Promotional Banner

Similar Questions

Explore conceptually related problems

The vapour pressure of pure water at 75^(@) is 296 torr the vapour pressure lowering due to 0.1 m solute is

6.0 g of urea (molecular weight=60) was dissolved in 9.9 moles of water. If the vapour pressure of pure water is P^@ , the vapour pressure of solution is :

What weight of solute (mol. Wt. 60) is required to dissolve in 180 g of water to reduce the vapour pressure to 4//5^(th) of pure water ?

What weight of solute (mol. Wt. 60) is required to dissolve in 180 g of water to reduce the vapour pressure to 4//5^(th) of pure water ?

What weight of solute (mol. Wt. 60) is required to dissolve in 180 g of water to reduce the vapour pressure to 4//5^(th) of pure water ?

6.0 g of urea (molecular mass = 60) was dissolved in 9.9 moles of water. If the vapour pressure of pure water is P^(@) , the vapour pressure of solution is :

6g of urea ( molecular weight =60) was dissolved in 9.9 moles of water. If the vapour pressure of pure water is P^(@) , the vapour pressure of solution is :