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(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 $25$.

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