A solution containing 4 g of polymer in 4.0 | vedclass

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

Question

(B) The osmotic pressure formula is $\pi = CRT$, where $C$ is the molarity, $R$ is the gas constant $(0.0821 \ L \ atm \ K^{-1} \ mol^{-1})$, and $T$

Vedclass · Accepted Answer

$82000$

Vedclass · Answer

(B) The osmotic pressure formula is $\pi = CRT$, where $C$ is the molarity, $R$ is the gas constant $(0.0821 \ L \ atm \ K^{-1} \ mol^{-1})$, and $T$ is the temperature in Kelvin.Given: $\pi = 3.0 	imes 10^{-4} \ atm$, $T = 27 + 273 = 300 \ K$, $V = 4.0 \ L$, and mass of solute $w = 4 \ g$.$C = \frac{\pi}{RT} = \frac{3.0 	imes 10^{-4}}{0.0821 	imes 300} = 1.218 	imes 10^{-5} \ mol \ L^{-1}$.Since $C = \frac{w}{M 	imes V}$, where $M$ is the molar mass:$1.218 	imes 10^{-5} = \frac{4}{M 	imes 4.0}$.$M = \frac{4}{1.218 	imes 10^{-5} 	imes 4.0} = \frac{1}{1.218 	imes 10^{-5}} \approx 82101 \ g \ mol^{-1}$.Rounding to the nearest given option, the molar mass is $82000 \ g \ mol^{-1}$.

$A$ solution containing $4 \ g$ of polymer in $4.0 \ L$ solution at $27^{\circ} C$ shows an osmotic pressure of $3.0 \times 10^{-4} \ atm$. The molar mass of the polymer in $g \ mol^{-1}$ is

Similar Questions

The osmotic pressure of a $5\%$ solution of cane sugar at $150\,^{\circ}C$ is ......... $atm$. (mol. wt. of cane sugar $= 342$)

$A$ $0.002 \, M$ solution of $NaCl$ having a degree of dissociation of $90 \%$ at $27 \, ^{\circ}C$ has an osmotic pressure equal to:

The empirical formula of a non-electrolyte is $CH_{2}O$. $A$ solution containing $6 \ g$ of the compound exerts the same osmotic pressure as that of $0.05 \ M$ glucose solution at the same temperature. The molecular formula of the compound is

Isotonic solutions are solutions having the same

Which method is used to remove salt from sea water?

Vedclass Test Series

Exam Paper Generator

Online Exam Module