When 10 g of glucose (P_1),10 g of urea (P_ | vedclass

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

Question

(C) Osmotic pressure $(\Pi)$ is given by the formula $\Pi = CRT = \frac{n}{V}RT = \frac{w}{M 	imes V}RT$. Since $w$ (mass),$V$ (volume),$R$ (gas cons

Vedclass · Accepted Answer

$P_2 > P_1 > P_3$

Vedclass · Answer

(C) Osmotic pressure $(\Pi)$ is given by the formula $\Pi = CRT = \frac{n}{V}RT = \frac{w}{M 	imes V}RT$. Since $w$ (mass),$V$ (volume),$R$ (gas constant),and $T$ (temperature) are constant for all three solutions,the osmotic pressure is inversely proportional to the molar mass $(M)$ of the solute: $\Pi \propto \frac{1}{M}$. The molar masses are: Urea $(NH_2CONH_2)$ = $60 \ g/mol$,Glucose $(C_6H_{12}O_6)$ = $180 \ g/mol$,and Sucrose $(C_{12}H_{22}O_{11})$ = $342 \ g/mol$. Since $60  P_1 > P_3$.

When $10 \ g$ of glucose $(P_1)$,$10 \ g$ of urea $(P_2)$,and $10 \ g$ of sucrose $(P_3)$ are dissolved in $250 \ mL$ of water,what is the relationship between their osmotic pressures at $273 \ K$?

Similar Questions

State the formula for osmotic pressure and provide examples.

$A$ solution containing $10 \, g \, dm^{-3}$ of urea $(M.wt = 60)$ is isotonic with a $5 \%$ solution of a non-volatile solute. The molecular mass of this non-volatile solute is ......... $g \, mol^{-1}$.

Which method is most suitable to determine the molecular mass of a polymer?

$A$ solution is made by dissolving $20 \ g$ of a substance in $500 \ mL$ of water. Its osmotic pressure was found to be $600 \ mm$ of $Hg$ at $15 \ ^oC$. The molecular weight of the substance is

Which one of the following pairs of solutions will be isotonic at the same temperature?

Vedclass Test Series

Exam Paper Generator

Online Exam Module