At 300 K,the osmotic pressure of a solution | vedclass

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(A) The osmotic pressure $\pi$ is given by the formula $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the tempera

Vedclass · Accepted Answer

$0.061$

Vedclass · Answer

(A) The osmotic pressure $\pi$ is given by the formula $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.For the first solution: $\pi_1 = C_1 RT = 4.98 \ bar$.The concentration $C_1 = \frac{n}{V} = \frac{36 \ g / 180 \ g/mol}{1 \ L} = 0.2 \ M$.Thus,$4.98 = 0.2 	imes R 	imes 300$.For the second solution at the same temperature: $\pi_2 = C_2 RT = 1.52 \ bar$.Dividing the two equations: $\frac{\pi_2}{\pi_1} = \frac{C_2}{C_1}$.$\frac{1.52}{4.98} = \frac{C_2}{0.2}$.$C_2 = \frac{1.52 	imes 0.2}{4.98} \approx 0.061 \ M$.

At $300 \ K$,the osmotic pressure of a solution containing $36 \ g$ of glucose per liter is $4.98 \ bar$. If the osmotic pressure of the solution is $1.52 \ bar$ at the same temperature,what will be its concentration (in $M$)?

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