At 300 K,a one litre solution of sucrose (mo | vedclass

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Question

(A) The formula for osmotic pressure is $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.

Vedclass · Accepted Answer

$292$

Vedclass · Answer

(A) The formula for osmotic pressure is $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.First,calculate the molar concentration $C$:$C = \frac{	ext{moles of solute}}{	ext{volume of solution in litres}} = \frac{40 \ g / 342 \ g \ mol^{-1}}{1 \ L} = 0.11696 \ mol \ L^{-1}$.Given $R = 8.314 	imes 10^6 \ cm^3 \ Pa \ K^{-1} \ mol^{-1}$. Since $1 \ L = 1000 \ cm^3$,$R = 8.314 	imes 10^3 \ L \ Pa \ K^{-1} \ mol^{-1}$.Now,calculate $\pi$:$\pi = 0.11696 \ mol \ L^{-1} 	imes 8.314 	imes 10^3 \ L \ Pa \ K^{-1} \ mol^{-1} 	imes 300 \ K$.$\pi = 291.71 	imes 10^3 \ Pa = 291.71 \ kPa$.Rounding to the nearest whole number,$\pi \approx 292 \ kPa$.

At $300 \ K$,a one litre solution of sucrose (molecular weight: $342$) was prepared by dissolving $40 \ g$ of sucrose. What is the approximate osmotic pressure (in $kPa$) of the solution at the same temperature? $(R = 8.314 \times 10^6 \ cm^3 \ Pa \ K^{-1} \ mol^{-1})$

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