At 300 K,the osmotic pressure of a solution | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The osmotic pressure $\pi$ is given by the formula $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the tempera

Vedclass · Accepted Answer

$0.061$

Vedclass · Answer

(A) The osmotic pressure $\pi$ is given by the formula $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.For the first solution: $\pi_1 = C_1 RT = 4.98 \ bar$.The concentration $C_1 = \frac{n}{V} = \frac{36 \ g / 180 \ g/mol}{1 \ L} = 0.2 \ M$.Thus,$4.98 = 0.2 	imes R 	imes 300$.For the second solution at the same temperature: $\pi_2 = C_2 RT = 1.52 \ bar$.Dividing the two equations: $\frac{\pi_2}{\pi_1} = \frac{C_2}{C_1}$.$\frac{1.52}{4.98} = \frac{C_2}{0.2}$.$C_2 = \frac{1.52 	imes 0.2}{4.98} \approx 0.061 \ M$.

At $300 \ K$,the osmotic pressure of a solution containing $36 \ g$ of glucose per liter is $4.98 \ bar$. If the osmotic pressure of the solution is $1.52 \ bar$ at the same temperature,what will be its concentration (in $M$)?

Similar Questions

What is the effect of external pressure on the osmotic pressure $(OP)$ of a solution?

Calculate the molar mass of a solute when $4 \ g$ of it is dissolved in $1 \ dm^3$ of solvent,having an osmotic pressure of $2 \ atm$ at $300 \ K$. $[R = 0.082 \ dm^3 \ atm \ K^{-1} \ mol^{-1}]$

Which of the following statements regarding isotonic solutions is $NOT$ correct?

Calculate the osmotic pressure exerted by a solution containing $0.822 \ g$ of solute in $300 \ mL$ of water at $300 \ K$. (Molar mass of solute $= 340 \ g \ mol^{-1}, R = 0.0821 \ L \ atm \ mol^{-1} \ K^{-1}$) (in $atm$)

An aqueous solution of $0.6 \% (w/v)$ urea is isotonic with which of the following?

Vedclass Test Series

Exam Paper Generator

Online Exam Module