At 27,^oC,the osmotic pressure of a solution | vedclass

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(B) Given,$\pi_1 = 4.92\,atm$,$\pi_2 = 1.5\,atm$. For glucose $(C_6H_{12}O_6)$,molar mass $M = 180\,g/mol$. Concentration $C_1 = \frac{36\,g}{180\,g/m

Vedclass · Accepted Answer

$0.061$

Vedclass · Answer

(B) Given,$\pi_1 = 4.92\,atm$,$\pi_2 = 1.5\,atm$. For glucose $(C_6H_{12}O_6)$,molar mass $M = 180\,g/mol$. Concentration $C_1 = \frac{36\,g}{180\,g/mol 	imes 1\,L} = 0.2\,M$. Since osmotic pressure $\pi = CRT$,at constant temperature $T$,$\pi \propto C$. Therefore,$\frac{\pi_1}{\pi_2} = \frac{C_1}{C_2}$. $\frac{4.92}{1.5} = \frac{0.2}{C_2}$. $C_2 = \frac{0.2 	imes 1.5}{4.92} = \frac{0.3}{4.92} \approx 0.061\,mol/L$.

At $27\,^oC$,the osmotic pressure of a solution containing $36\,g$ of glucose per liter is $4.92\,atm$. If the osmotic pressure of the solution is adjusted to $1.5\,atm$ at the same temperature,what will be its concentration?

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