A solution of polystyrene in benzene contains | vedclass

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

Question

(B) The osmotic pressure $\pi$ is given by $\pi = h \cdot d \cdot g$,where $h$ is the height of the solution column,$d$ is the density,and $g$ is the

Vedclass · Accepted Answer

$25.2 	imes 10^3 \, g/mol$

Vedclass · Answer

(B) The osmotic pressure $\pi$ is given by $\pi = h \cdot d \cdot g$,where $h$ is the height of the solution column,$d$ is the density,and $g$ is the acceleration due to gravity.Given: $h = 11 \, cm = 0.11 \, m$,$d = 0.9 \, g/mL = 900 \, kg/m^3$,$C = 10 \, g/L = 10/M \, mol/L = 10/M 	imes 10^3 \, mol/m^3$,$T = 27 + 273 = 300 \, K$,$R = 8.314 \, J/(mol \cdot K)$.Using $\pi = CRT$:$h \cdot d \cdot g = (C/M) \cdot R \cdot T$$0.11 	imes 900 	imes 9.8 = (10/M) 	imes 10^3 	imes 8.314 	imes 300$$970.2 = (24942000 / M)$$M = 24942000 / 970.2 \approx 25708 \, g/mol$.Using the standard approximation $\pi = h \cdot d \cdot g$ and converting units to $atm$:$\pi = (11 \, cm 	imes 0.9 \, g/mL) / (13.6 \, g/mL 	imes 76 \, cm) \approx 0.00958 \, atm$.Using $\pi = CRT$ with $C = 10/M \, g/L$:$0.00958 = (10/M) 	imes 0.0821 	imes 300$$M = (10 	imes 0.0821 	imes 300) / 0.00958 \approx 25.2 	imes 10^3 \, g/mol$.

$A$ solution of polystyrene in benzene contains $10 \, g/L$. The equilibrium height of the column of solution having density $0.9 \, g/mL$ in the osmometer is $11 \, cm$ at $27 \, ^\circ C$. What is the average molar mass of polystyrene? (Assume the solution is ideal)

Similar Questions

$A$ solution in which red blood cells can remain in their normal form is called ......

$.......... \%$ percent of glucose solution is isotonic with $1.5 \% \, (w/V)$ urea solution.

$A$ solution contains a non-volatile solute of molecular mass $M_p$. Which of the following can be used to calculate the molecular mass of the solute in terms of osmotic pressure ($m = \text{Mass of solute}$,$V = \text{Volume of solution}$,and $\pi = \text{Osmotic pressure}$)?

$A$ solution of a nonvolatile solute is obtained by dissolving $0.8 \ g$ in $0.3 \ dm^3$ of water. The solution has an osmotic pressure of $0.2 \ atm$ at $300 \ K$. Calculate the molar mass of the solute. $[R = 0.082 \ atm \ dm^3 \ K^{-1} \ mol^{-1}]$

$18 \ g$ glucose (Molar mass $= 180 \ g/mol$) is dissolved in $100 \ mL$ of water at $300 \ K$. If $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,what is the osmotic pressure of the solution (in $atm$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module