The osmotic pressure of a 5% (wt/vol) solutio | vedclass

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(B) The osmotic pressure $\pi$ is given by the formula $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant $(0.082 \ L \ atm \ K^

Vedclass · Accepted Answer

$5.08$

Vedclass · Answer

(B) The osmotic pressure $\pi$ is given by the formula $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant $(0.082 \ L \ atm \ K^{-1} \ mol^{-1})$,and $T$ is the temperature in Kelvin.$1.$ Calculate the molar concentration $(C)$: $A$ $5\%$ (wt/vol) solution means $5 \ g$ of cane sugar ($C_{12}H_{22}O_{11}$,molar mass $= 342 \ g/mol$) in $100 \ mL$ of solution.$C = \frac{5 \ g}{342 \ g/mol} 	imes \frac{1000 \ mL/L}{100 \ mL} = \frac{50}{342} \ mol/L \approx 0.1462 \ mol/L$.$2.$ Convert temperature to Kelvin: $T = 150 + 273 = 423 \ K$.$3.$ Calculate osmotic pressure: $\pi = (\frac{50}{342} \ mol/L) 	imes (0.082 \ L \ atm \ K^{-1} \ mol^{-1}) 	imes (423 \ K) \approx 5.07 \ atm$.Thus,the closest value is $5.08 \ atm$.

The osmotic pressure of a $5\%$ (wt/vol) solution of cane sugar at $150\,^{\circ}C$ is $......... \ atm$.

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