The osmotic pressure of a living cell is 12 | vedclass

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Question

(A) For an isotonic solution,the osmotic pressure $(\pi)$ of the $NaCl$ solution must be equal to the osmotic pressure of the living cell.
$\pi = iCR

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) For an isotonic solution,the osmotic pressure $(\pi)$ of the $NaCl$ solution must be equal to the osmotic pressure of the living cell.
$\pi = iCRT$
Given $\pi = 12 \ atm$,$T = 300 \ K$,$R = 0.08 \ L \ atm \ K^{-1} \ mol^{-1}$,and $i = 2$ (for $NaCl ightarrow Na^{+} + Cl^{-}$).
$12 = 2 	imes C 	imes 0.08 	imes 300$
$12 = 48 	imes C$
$C = \frac{12}{48} = 0.25 \ mol \ L^{-1}$
The molar mass of $NaCl = 23 + 35.5 = 58.5 \ g \ mol^{-1}$.
Strength of $NaCl$ solution = $C 	imes 	ext{Molar mass}$
$= 0.25 	imes 58.5 = 14.625 \ g \ L^{-1}$.
Rounding to the nearest integer,we get $15 \ g \ L^{-1}$.

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