Home
Class 12
CHEMISTRY
An aqueous solution is prepared by disso...

An aqueous solution is prepared by dissolving 0.1 mol of an ionic salt in 1.8 kg of water at `35^(@)C`. The salt remains 90% dissociated in the solution. The vapour pressure of the solution is 59.724 mm of Hg. Vapor pressure of water at `35^(@)C` is 60.000 mm of Hg. The number of ions present per formula unit of the ionic salt is _______.

Text Solution

AI Generated Solution

Promotional Banner

Topper's Solved these Questions

  • JEE ADVANCED 2022

    JEE ADVANCED PREVIOUS YEAR|Exercise CHEMISTRY (SECTION 2)|6 Videos

  • JEE ADVANCED 2022

    JEE ADVANCED PREVIOUS YEAR|Exercise CHEMISTRY (SECTION 3)|4 Videos

  • JEE ADVANCED 2021

    JEE ADVANCED PREVIOUS YEAR|Exercise QUESTION|38 Videos

Similar Questions

Explore conceptually related problems

10 gram of a solute was dissolved in 80 g of acetone at 30ºC to form an ideal solution. The vapour pressure was found to be 271 mm of Hg. Vapour pressure of pure acetone at 30^(@) C is 283 mm Hg. The molecular weight of the solute is

The Vapour pressure of solution containing 2g of NaCl in 100g of water at 100^(@)C is 753.2mm of Hg , then degree of dissociation of NaCl .

The vapour pressure of 2.1% solution of a non- electrolyte in water at 100^(@) C is 755 mm Hg. Calculate the molar mass of the solute .

A solution is prepared by dissolving 10g of non-volatile solute in 200g of water. It has a vapour pressure of 31.84 mm Hg at 308 K. Calculate the molar mass of the solute. (Vapour pressure of pure water at 308K =32 mm Hg)

When 25g of a non-volatile solute is dissolved in 100g of water, the vapour pressure is lowered by 0.225 mm. If the vapour pressure of water at 25^(@)C is 17.5 mm, what is the molecular mass of solute ?