Home
Class 12
CHEMISTRY
A solution is prepared by dissolving 0.6...

A solution is prepared by dissolving 0.6 g of urea (molar mass `=60 g"mol"^(-1)`) and 1.8 g of glucose (molar mass `=180g"mol"^(-1)`) in 100 mL of water at `27^(@)C`. The osmotic pressure of the solution is:
`(R=0.8206L "atm"K^(-1)"mol"^(-1))`

A

1.64 atm

B

4.92 atm

C

8.2 atm

D

2.46 atm

Text Solution

Verified by Experts

`pi=pi_(1)+pi_(2)`
`pi_(1)=` osmotic pressure of urea solution.
`pi_(2)=` osmotic pressure of glucose solution.
`pi_(1)=C_(1)RT" "pi_(2)=C_(2)RT`
`C_(1)=(0.6//60)/0.1=0.1M`
`C_(2)=(1.8//180)/0.1=0.1M`
`pi=pi_(1)+pi_(2)=0.1RT+0.1RT`
`pi-2(0.1RT)=0.2xx24.6=4.92atm`.
Promotional Banner

Similar Questions

Explore conceptually related problems

If 23 g of glucose ( molecular mass 180)is dissolved in 60ml of water at 15^(@)C , then the osmotic pressure of this solution will be

9g of glucose and 3g of urea are dissolved in 1L of solution at 27^(@)C .The osmotic pressure of the solution will be

36g of glucose (molar mass = 180 g//mol) is present in 500g of water, the molarity of the solution is

The molarity of urea (molar mass 60 g mol^(-1) ) solution by dissolving 15 g of urea in 500 cm^(3) of water is

The density of a solution prepared by dissolving 120 g of urea (mol. Mass=60 u) in 1000 g of water is 1.15 g/mL. The molarity if this solution is