The solution containing 4.0 g of a polyvinyl | vedclass

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

Question

(B) The formula for osmotic pressure is $\pi V = nRT = \frac{w}{M} RT$. Given: $\pi = 6.0 	imes 10^{-4} \ atm$,$V = 1 \ L$,$w = 4.0 \ g$,$T = 300 \ K

Vedclass · Accepted Answer

$1.6 	imes 10^{5}$

Vedclass · Answer

(B) The formula for osmotic pressure is $\pi V = nRT = \frac{w}{M} RT$. Given: $\pi = 6.0 	imes 10^{-4} \ atm$,$V = 1 \ L$,$w = 4.0 \ g$,$T = 300 \ K$,and $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$. Substituting the values: $6.0 	imes 10^{-4} 	imes 1 = \frac{4.0}{M} 	imes 0.082 	imes 300$. $M = \frac{4.0 	imes 0.082 	imes 300}{6.0 	imes 10^{-4}}$. $M = \frac{98.4}{6.0 	imes 10^{-4}} = 16.4 	imes 10^{4} = 1.64 	imes 10^{5} \ g \ mol^{-1}$. Thus,the molecular mass is approximately $1.6 	imes 10^{5} \ g \ mol^{-1}$.

The solution containing $4.0 \ g$ of a polyvinyl chloride polymer in $1 \ L$ of dioxane was found to have an osmotic pressure of $6.0 \times 10^{-4} \ atm$ at $300 \ K$. Given $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$,the molecular mass of the polymer is:

Similar Questions

At $27^{\circ}C$,a solution containing $2.5 \ g$ of solute in $250.0 \ mL$ of solution exerts an osmotic pressure of $400 \ Pa$. The molar mass of the solute is $.............. \ g \ mol^{-1}$ (Nearest integer) (Given: $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$)

Which of the following has the highest osmotic pressure at $300 \, K$?

What would happen if a thin slice of sugar beet is placed in a concentrated solution of $NaCl$?

$A$ semipermeable membrane used in the measurement of osmotic pressure of a solution allows the passage of

Assertion : If red blood cells were removed from the body and placed in pure water,pressure inside the cells increases.
Reason : The concentration of salt content in the cells increases.

Vedclass Test Series

Exam Paper Generator

Online Exam Module