If 3 g of glucose (mol. wt. 180) is dissolve | vedclass

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(C) The formula for osmotic pressure is $\pi = CRT$,where $C$ is the molarity,$R$ is the gas constant $(0.0821 \ L \ atm \ K^{-1} \ mol^{-1})$,and $T$

Vedclass · Accepted Answer

$6.57$

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(C) The formula for osmotic pressure is $\pi = CRT$,where $C$ is the molarity,$R$ is the gas constant $(0.0821 \ L \ atm \ K^{-1} \ mol^{-1})$,and $T$ is the temperature in Kelvin.First,calculate the number of moles of glucose: $n = \frac{3 \ g}{180 \ g/mol} = 0.01667 \ mol$.The volume of the solution is approximately equal to the volume of the solvent (water) since the mass of solute is small. Assuming density of water $\approx 1 \ g/mL$,$V = 60 \ mL = 0.06 \ L$.Molarity $C = \frac{n}{V} = \frac{0.01667 \ mol}{0.06 \ L} = 0.2778 \ M$.Temperature $T = 15 + 273 = 288 \ K$.Osmotic pressure $\pi = 0.2778 	imes 0.0821 	imes 288 = 6.568 \ atm \approx 6.57 \ atm$.

If $3 \ g$ of glucose (mol. wt. $180$) is dissolved in $60 \ g$ of water at $15 \ ^oC$. Then the osmotic pressure of this solution will be .......... $atm$.

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