Calculate the temperature of 0.05 M sucrose | vedclass

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Question

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

Vedclass · Accepted Answer

$365.4$

Vedclass · Answer

(D) The formula for osmotic pressure $(\pi)$ is given by $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin. Given values: $\pi = 1.5 \ atm$,$C = 0.05 \ M$,$R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$. Rearranging the formula to solve for $T$: $T = \frac{\pi}{CR}$. Substituting the values: $T = \frac{1.5}{0.05 	imes 0.0821} = \frac{1.5}{0.004105} \approx 365.4 \ K$. Therefore,the correct option is $D$.

Calculate the temperature of $0.05 \ M$ sucrose solution in Kelvin if the osmotic pressure of the solution is $1.5 \ atm$. $[R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}]$

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