Question

Osmotic pressure of $0.01M$ aqueous urea is $0.24atm.$ If $0.02M$ of $CaC{l}_2\left(aq\right)$ is taken at same temperature, then $[\text{Take : R}=0.08atmLmo{l}^{-1}{K}^{-1}]$

• A. Temperature of urea solution is $300K$
• B. Osmotic pressure of $CaC{l}_2\left(aq\right)$ is $1.44atm$
• C. If $100mL$ of both solution are mixed at $300K$, then osmotic pressure of resultant mixture will be $0.84atm$
• D. Vapor presesure of $0.01M$ aqueous urea will be greater than $0.02MCaC{l}_2\left(aq\right)$

Answer 1

Answer: (A), (B), (C), (D)

Vapor presesure of $0.01M$ aqueous urea will be greater than $0.02MCaC{l}_2\left(aq\right)$

(A) Osmotic pressure $\left(\pi \right)=iCRT(i=1\text{for urea})$
$\Rightarrow 0.24=0.01\times 0.08\times T$
$\Rightarrow T=\frac{0.24}{0.01\times 0.08}=300K$

(B) Osmotic pressure of $CaC{l}_2=iCRT(i=3\text{for}CaC{l}_2)$
$=3\times 0.02\times 0.08\times 300$
$=1.44atm$

(C) The resultanr concentration of the solution obtained by mixing 100 ml of each of the solution is,
$\frac{0.01\times 100+3\times 0.02\times 100}{200}=0.035mol/L$
So the osmotic pressure will be,
$\pi =0.035\times 0.08\times 300=0.84atm$

(D) Since $CaC{L}_2$ will disssociate to give ions, thus number od solute particles will be more in case of $CaC{l}_2$ and accordingly vapour pressure will be less for $0.02MCaC{l}_2$ compared to $0.01M$ urea.