If 3 g of glucose (molar mass = 180 g mol^ | vedclass

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(B) Given: Molar mass of glucose $(M_B) = 180 \ g \ mol^{-1}$.Mass of glucose $(W_B) = 3 \ g$.Mass of water $(W_A) = 60 \ g$.Temperature $(T) = 15^{\c

Vedclass · Accepted Answer

$6.57$

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(B) Given: Molar mass of glucose $(M_B) = 180 \ g \ mol^{-1}$.Mass of glucose $(W_B) = 3 \ g$.Mass of water $(W_A) = 60 \ g$.Temperature $(T) = 15^{\circ} C = 273 + 15 = 288 \ K$.Assuming the density of the solution is approximately $1 \ g \ mL^{-1}$,the volume of the solution $(V)$ is approximately $60 \ mL = 0.06 \ L$.Number of moles of glucose $(n_B) = \frac{W_B}{M_B} = \frac{3}{180} = 0.01667 \ mol$.Molarity $(C) = \frac{n_B}{V(L)} = \frac{0.01667}{0.06} = 0.2778 \ mol \ L^{-1}$.Using the formula for osmotic pressure: $\pi = CRT$.$\pi = 0.2778 	imes 0.0821 	imes 288 = 6.568 \ atm \approx 6.57 \ atm$.

If $3 \ g$ of glucose (molar mass $= 180 \ g \ mol^{-1}$) is dissolved in $60 \ g$ of water at $15^{\circ} C$,the osmotic pressure of the solution will be (in $atm$)

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