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If $3 \mathrm{~g}$ of glucose (molar mass $=180 \mathrm{~g})$ is dissolved in $60 \mathrm{~g}$ of water at $15^{\circ} \mathrm{C}$, the osmotic pressure of the solution will be
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The correct answer is:
$6.57 \mathrm{~atm}$
Given, molar mass of glucose $M_B=180 \mathrm{~g}$
Mass of glucose, $W_B=3 \mathrm{~g}$
Mass of water, $W_A=60 \mathrm{~g}$
Temperature $=15^{\circ} \mathrm{C} \Rightarrow 273+15=288 \mathrm{~K}$
Osmotic pressure, $\pi=$ ?
$$
C=\frac{W_B \times 1000}{W_A \times M_B}=\frac{3 \times 1000}{60 \times 180}=0.277 \mathrm{~mol} \mathrm{~L}^{-1}
$$
We know that,
$$
\begin{aligned}
\pi & =C R T=0.277 \times 0.0821 \times 288 \\
& =6.549 \simeq 6.57 \mathrm{~atm}
\end{aligned}
$$
Mass of glucose, $W_B=3 \mathrm{~g}$
Mass of water, $W_A=60 \mathrm{~g}$
Temperature $=15^{\circ} \mathrm{C} \Rightarrow 273+15=288 \mathrm{~K}$
Osmotic pressure, $\pi=$ ?
$$
C=\frac{W_B \times 1000}{W_A \times M_B}=\frac{3 \times 1000}{60 \times 180}=0.277 \mathrm{~mol} \mathrm{~L}^{-1}
$$
We know that,
$$
\begin{aligned}
\pi & =C R T=0.277 \times 0.0821 \times 288 \\
& =6.549 \simeq 6.57 \mathrm{~atm}
\end{aligned}
$$
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