Home
Class 12
CHEMISTRY
The vapour pressure of C Cl4 at 25^(@)C ...

The vapour pressure of `C Cl_4` at `25^(@)C` is 143 mm of Hg. 0.5 gm of a non-volatile solute (mol. wt. 65) is dissolved in 100ml of `C Cl_4` , the vapour pressure of the solution will be (Density of `C Cl_4`=1.58g/`cm^(3)` )

Text Solution

AI Generated Solution

To solve the problem, we will use Raoult's Law, which states that the vapor pressure of a solution (P) is related to the vapor pressure of the pure solvent (P₀) and the mole fraction of the solvent in the solution. ### Step-by-Step Solution: **Step 1: Calculate the number of moles of the solute.** - Given: - Mass of solute (W) = 0.5 g - Molar mass of solute (M) = 65 g/mol ...
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGMENT (SECTION-A)|50 Videos

  • SOLUTIONS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGMENT (SECTION-B)|21 Videos

  • SOLUTIONS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGMENT (SECTION-J) AAKASH CHALLENGERS QUESTIONS|10 Videos

  • REDOX REACTIONS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION - D|10 Videos

  • SOME BASIC CONCEPT OF CHEMISTRY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT( SECTION - D) Assertion-Reason Type Questions|15 Videos

Similar Questions

Explore conceptually related problems

The vapour pressure of C Cl_4 at 25^@C is 143 mm Hg. If 0.5 gm of a non-volatile solute (mol.weight=65) is dissolved in 100g C Cl_4 , the vapour pressure of the solution will be

Vapour pressure of C CL_(4) at 25^@C is 143 mmHg 0.05g of a non-volatile solute (mol.wt.= 65 )is dissolved in 100ml C CL_(4) . find the vapour pressure of the solution (density of C CL_(4)=158g//cm^2 )

The vapour pressure of water at 25^(@)C is 18 mmHg. If 18 g of glucose (C_(6)H_(12)O_(6)) is added to 178.2 g of water at 20^(@)C , the vapour pressure of the resulting solution will be :

The vapour pressure of water at 20^(@) C is 17.5 mm Hg. If 18 g of gulucose (C_(6)H_(12)O_(6)) is added to 178.2 g of water at 20^(@) C, the vapour pressure of the resulting solution will be:

The vapour pressure of water at 20^(@)C is 17.54mm . When 20g of non - ionic substance is dissolved in 100g of water, the vapour pressure is lowered by 0.30mm . What is the molecular mass of the substance ?

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 ?

The vapour pressure of acetone at 20^(@)C is 185 torr. When 1.2 g of non-volatile substance was dissolved in 100g of acetone at 20^(@)C its vapour pressure was 183 torr. The molar mass (g mol^(-1)) of the substance is:

Vapour pressure of benzene at 30^(@)C is 121.8 mm. When 15 g of a non-volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm. The molecular weight of the solute is (mol. Weight of solvent = 78)

The vapour pressure of benzene at 30^(@)C is 164.33 mm of Hg . In 3 mol of benzene, when 6 g of acetic acid was dissolved, the vapour pressure of the solution became 162.04 mm of Hg . Calculate a. Van't Hoff factor b.The degree fo association of acetic acid in bezene at 30^(@)C .