Calculate the osmotic pressure exerted by a s | vedclass

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

Question

(B) The formula for osmotic pressure is $\pi = CRT$. First,calculate the number of moles of solute: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac

Vedclass · Accepted Answer

$0.2$

Vedclass · Answer

(B) The formula for osmotic pressure is $\pi = CRT$. First,calculate the number of moles of solute: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{0.822 \ g}{340 \ g \ mol^{-1}} \approx 0.0024176 \ mol$. Next,calculate the molar concentration $C$ in $mol \ L^{-1}$: $C = \frac{n}{V(L)} = \frac{0.0024176 \ mol}{0.300 \ L} \approx 0.0080588 \ M$. Now,substitute the values into the osmotic pressure equation: $\pi = 0.0080588 \ mol \ L^{-1} 	imes 0.0821 \ L \ atm \ mol^{-1} \ K^{-1} 	imes 300 \ K$. $\pi \approx 0.1985 \ atm \approx 0.2 \ atm$.

Calculate the osmotic pressure exerted by a solution containing $0.822 \ g$ of solute in $300 \ mL$ of water at $300 \ K$. (Molar mass of solute $= 340 \ g \ mol^{-1}, R = 0.0821 \ L \ atm \ mol^{-1} \ K^{-1}$) (in $atm$)

Similar Questions

The osmotic pressure of a solution containing $34.2 \ g$ of cane sugar (molar mass = $342 \ g \ mol^{-1}$) in $1 \ L$ of solution at $20^{\circ} C$ is
(Given,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$) (in $atm$)

Which of the following $0.1 \ M$ solutions has the maximum osmotic pressure at $25^o \ C$?

The pressure required for the reverse osmosis is quite high. For this,which of the following membranes is used?

Which of the following methods is used to obtain pure water from seawater?

Which will have minimum osmotic pressure?

Vedclass Test Series

Exam Paper Generator

Online Exam Module