The osmotic pressure of a solution containing | vedclass

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(C) The formula for osmotic pressure is $\pi = CRT = \frac{n}{V} RT$. Given: Mass of glucose $(w_2)$ = $3 \ g$,Molar mass of glucose $(M_2)$ = $180 \

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$6.25$

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(C) The formula for osmotic pressure is $\pi = CRT = \frac{n}{V} RT$. Given: Mass of glucose $(w_2)$ = $3 \ g$,Molar mass of glucose $(M_2)$ = $180 \ g/mol$,Moles of glucose $(n_2)$ = $\frac{3}{180} = 0.01667 \ mol$. Mass of water $(w_1)$ = $60 \ g$,Density of solution $(ho)$ = $1 \ g/mL$. Since the amount of solute is very small,the volume of the solution $(V)$ is approximately equal to the volume of water = $60 \ mL = 0.06 \ L$. Temperature $(T)$ = $15 + 273.15 = 288.15 \ K$. Gas constant $(R)$ = $0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1}$. $\pi = \frac{0.01667 \ mol}{0.06 \ L} 	imes 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} 	imes 288.15 \ K$. $\pi \approx 6.56 \ atm$. Given the options provided,the closest value is $6.25 \ atm$.

The osmotic pressure of a solution containing $3 \ g$ of glucose (molar mass = $180 \ g/mol$) in $60 \ g$ of water at $15^\circ C$ (density = $1 \ g/mL$) is ....... $atm$.

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