Calculate the molar mass of a non-electrolyte | vedclass

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(C) The formula for osmotic pressure $(\pi)$ is $\pi = CRT = \frac{n}{V} RT = \frac{w}{M \cdot V} RT$.Given: $\pi = 2.4 	ext{ atm}$, $w = 6 	ext{ g}

Vedclass · Accepted Answer

$61.58 	ext{ g mol}^{-1}$

Vedclass · Answer

(C) The formula for osmotic pressure $(\pi)$ is $\pi = CRT = \frac{n}{V} RT = \frac{w}{M \cdot V} RT$.Given: $\pi = 2.4 	ext{ atm}$, $w = 6 	ext{ g}$, $V = 1 	ext{ dm}^3$, $T = 300 	ext{ K}$, $R = 0.0821 	ext{ atm dm}^3 	ext{ K}^{-1} 	ext{ mol}^{-1}$.Rearranging for molar mass $(M)$: $M = \frac{wRT}{\pi V}$.Substituting the values: $M = \frac{6 	imes 0.0821 	imes 300}{2.4 	imes 1}$.$M = \frac{147.78}{2.4} = 61.575 	ext{ g mol}^{-1} \approx 61.58 	ext{ g mol}^{-1}$.

Calculate the molar mass of a non-electrolyte solute when $6 \text{ g}$ of it is dissolved in $1 \text{ dm}^3$ of water, having an osmotic pressure of $2.4 \text{ atm}$ at $300 \text{ K}$ $(R = 0.0821 \text{ atm dm}^3 \text{ K}^{-1} \text{ mol}^{-1})$.

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