A 40 text{ g} non-electrolyte solute having a | vedclass

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Question

(A) Step $1$: Calculate the number of moles of solute $(n)$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{40 	ext{ g}}{180 	ext{ g mol}^{-1}}

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(A) Step $1$: Calculate the number of moles of solute $(n)$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{40 	ext{ g}}{180 	ext{ g mol}^{-1}} = \frac{2}{9} 	ext{ mol} \approx 0.222 	ext{ mol}$.Step $2$: Use the osmotic pressure formula $\Pi V = nRT$, where $\Pi = 2 	ext{ atm}$, $n = \frac{2}{9} 	ext{ mol}$, $R = 0.0821 	ext{ L atm K}^{-1} 	ext{mol}^{-1}$, and $T = 300 	ext{ K}$.Step $3$: Rearrange to solve for $V$: $V = \frac{nRT}{\Pi} = \frac{(2/9) 	imes 0.0821 	imes 300}{2}$.Step $4$: Calculate the value: $V = \frac{0.2222 	imes 0.0821 	imes 300}{2} = \frac{5.472}{2} = 2.736 	ext{ L}$.Note: Since $1 	ext{ L} = 1 	ext{ dm}^3$, the volume is approximately $2.74 	ext{ dm}^3$. Given the options, there appears to be a discrepancy in the provided choices. Re-evaluating with $R \approx 0.082$ and $n = 40/180$: $V = (40/180 	imes 0.0821 	imes 300) / 2 = 2.736 	ext{ dm}^3$. None of the options match.

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