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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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}}

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(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.

Similar Questions

$0.2 \ M$ aqueous solution of glucose has osmotic pressure $4.9 \ atm$ at $300 \ K$. What is the concentration of glucose if it has osmotic pressure $1.5 \ atm$ at same temperature (in $M$)?

Explain: What is osmosis? Give examples.

The correct relationship between the osmotic pressure of $0.1 \ M \ NaCl$ and $0.1 \ M \ Na_2SO_4$ is ......

What is the relationship between the osmotic pressures of $6.00 \ g \ L^{-1}$ $CH_3COOH$ $(\pi_1)$ and $7.45 \ g \ L^{-1}$ $KCl$ $(\pi_2)$ solutions?

Which of the following aqueous solutions are isotonic $(R = 0.082 \ atm \ K^{-1} mol^{-1})$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module