An artificial cell is made by encapsulating 0 | vedclass

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

Question

(C) The osmotic pressure $\pi$ is given by the formula $\pi = \Delta C 	imes R 	imes T$,where $\Delta C$ is the difference in molar concentrations o

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) The osmotic pressure $\pi$ is given by the formula $\pi = \Delta C 	imes R 	imes T$,where $\Delta C$ is the difference in molar concentrations of the particles.For the glucose solution inside the cell: $C_{glucose} = 0.2 \ M$.For the $NaCl$ solution outside the cell,assuming complete dissociation $(NaCl ightarrow Na^{+} + Cl^{-})$,the total concentration of particles is $C_{NaCl} = 2 	imes 0.05 \ M = 0.1 \ M$.The net concentration difference is $\Delta C = C_{glucose} - C_{NaCl} = 0.2 \ M - 0.1 \ M = 0.1 \ M$.Now,calculate the osmotic pressure:$\pi = 0.1 	imes 0.083 	imes 300$$\pi = 2.49 \ 	ext{bar}$To express this in $	imes 10^{-1} \ 	ext{bar}$:$\pi = 24.9 	imes 10^{-1} \ 	ext{bar}$.Rounding to the nearest integer,we get $25$.

Similar Questions

Under identical conditions,which solution has the highest osmotic pressure?

Which of the following is a colligative property?

$A$ solution containing $8.6 \ g$ of urea per liter is isotonic with a $0.5 \ \%$ (weight/volume) solution of a non-volatile organic solute. What is the molecular weight of the organic solute?

The osmotic pressure of a sugar solution at $300 \ K$ is $2.46 \ atm$. What is the concentration of the solution (in $M$)?

Arrange the following solutions according to decreasing order of osmotic pressure under similar conditions of temperature,assuming complete dissociation:
$A. 0.2 \ m \ KCl$
$B. 0.3 \ m \ MgSO_4$
$C. 0.1 \ m \ BaCl_2$
$D. 0.5 \ m \ Al_2(SO_4)_3$

Vedclass Test Series

Exam Paper Generator

Online Exam Module