At 27^{circ} C,the osmotic pressure of a solu | vedclass

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

Question

(C) The formula for osmotic pressure $(\pi)$ is $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in

Vedclass · Accepted Answer

$249$

Vedclass · Answer

(C) The formula for osmotic pressure $(\pi)$ is $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.Given: $\pi = 0.4 \ bar$,$w = 4 \ g$,$V = 1.0 \ L$,$R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$,$T = 27 + 273 = 300 \ K$.The molar concentration $C = \frac{n}{V} = \frac{w}{M 	imes V}$,where $M$ is the molar mass.Substituting the values: $0.4 = \frac{4}{M 	imes 1.0} 	imes 0.083 	imes 300$.$0.4 = \frac{4 	imes 0.083 	imes 300}{M}$.$M = \frac{4 	imes 0.083 	imes 300}{0.4} = \frac{99.6}{0.4} = 249 \ g \ mol^{-1}$.

At $27^{\circ} C$,the osmotic pressure of a solution containing $4 \ g$ of a non-electrolyte solute in $1.0 \ L$ of solution is $0.4 \ bar$. The molar mass of the solute in $g \ mol^{-1}$ is :
$(R=0.083 \ L \ bar \ K^{-1} \ mol^{-1})$

Similar Questions

If a $0.1 \ M$ solution of glucose (mol. wt. $180$) and $0.1 \ M$ solution of urea (mol. wt. $60$) are placed on the two sides of a semipermeable membrane to equal heights,then it will be correct to say

To have an osmotic pressure equal to a solution containing $9.2 \ g$ of glucose per liter,how many grams of glucose must be dissolved in $0.5 \ L$ of solution?

Calculate the osmotic pressure of $0.5 \ M$ aqueous solution of a non-volatile solute at $300 \ K$ $\left[R = 0.0821 \ atm \ dm^3 \ K^{-1} \ mol^{-1}\right]$. (in $atm$)

The osmotic pressure of a solution at $327\,^oC$ and concentration $C$ is $P$. If the osmotic pressure of the same solution at concentration $C/2$ and temperature $427\,^oC$ is $2\, atm$,then the value of $P$ is:

At $300 \, K$,$36 \, g$ of glucose present in a litre of its solution has an osmotic pressure of $4.98 \, bar$. If the osmotic pressure of the solution is $1.52 \, bar$ at the same temperature,what would be its concentration?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

At $27^{\circ} C$,the osmotic pressure of a solution containing $4 \ g$ of a non-electrolyte solute in $1.0 \ L$ of solution is $0.4 \ bar$. The molar mass of the solute in $g \ mol^{-1}$ is :$(R=0.083 \ L \ bar \ K^{-1} \ mol^{-1})$