The osmotic pressure (in atm) of an aqueous s | vedclass

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(A) The osmotic pressure $\pi$ is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the total molarity,$R$ is the gas consta

Vedclass · Accepted Answer

$2.43$

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(A) The osmotic pressure $\pi$ is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the total molarity,$R$ is the gas constant,and $T$ is the temperature in Kelvin.First,calculate the van't Hoff factor for $NaCl$: $i = 1 + \alpha(n-1)$. Here $\alpha = 0.94$ and $n = 2$,so $i = 1 + 0.94(2-1) = 1.94$.The effective moles of $NaCl$ particles = $i 	imes 	ext{moles} = 1.94 	imes 0.01 = 0.0194 \ mol$.Glucose is a non-electrolyte,so its moles remain $0.03 \ mol$.Total moles of solute = $0.0194 + 0.03 = 0.0494 \ mol$.Volume of solution = $500 \ mL = 0.5 \ L$.Total molarity $C = \frac{0.0494 \ mol}{0.5 \ L} = 0.0988 \ M$.Temperature $T = 27 + 273 = 300 \ K$.Osmotic pressure $\pi = 0.0988 	imes 0.082 	imes 300 = 2.43048 \ atm \approx 2.43 \ atm$.

The osmotic pressure (in $atm$) of an aqueous solution containing $0.01 \ mol$ of $NaCl$ (degree of dissociation $0.94$) and $0.03 \ mol$ of glucose in $500 \ mL$ at $27^{\circ} C$ is $\left(R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}\right)$

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