If 3 g of glucose (mol. wt. 180) is dissolve | vedclass

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

Question

(C) The formula for osmotic pressure 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

$6.57$

Vedclass · Answer

(C) The formula for osmotic pressure 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.First,calculate the number of moles of glucose: $n = \frac{3 \ g}{180 \ g/mol} = 0.01667 \ mol$.The volume of the solution is approximately equal to the volume of the solvent (water) since the mass of solute is small. Assuming density of water $\approx 1 \ g/mL$,$V = 60 \ mL = 0.06 \ L$.Molarity $C = \frac{n}{V} = \frac{0.01667 \ mol}{0.06 \ L} = 0.2778 \ M$.Temperature $T = 15 + 273 = 288 \ K$.Osmotic pressure $\pi = 0.2778 	imes 0.0821 	imes 288 = 6.568 \ atm \approx 6.57 \ atm$.

If $3 \ g$ of glucose (mol. wt. $180$) is dissolved in $60 \ g$ of water at $15 \ ^oC$. Then the osmotic pressure of this solution will be .......... $atm$.

Similar Questions

Which of the following solutions does not flow in either direction when separated by a semipermeable membrane? (Molar mass: glucose $= 180 \ g \ mol^{-1}$,urea $= 60 \ g \ mol^{-1}$)

Which of the following solutions will not show flow of solvent in either direction when separated by a semipermeable membrane?

The osmotic pressure of a dilute solution is directly proportional to the

Which of the following would have the highest osmotic pressure? (Assume that all salts are $90\%$ dissociated.)

$0.5 \ N$ solution of $KCl$ is isotonic with $.........$ solution of sugar,assuming $100\%$ dissociation for $KCl$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module