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KCET · Chemistry · Solutions

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

  1. A \(0.65 \mathrm{~atm}\)
  2. B \(6.57 \mathrm{~atm}\)
  3. C \(5.57 \mathrm{~atm}\)
  4. D \(0.34 \mathrm{~atm}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(6.57 \mathrm{~atm}\)

Step-by-step Solution

Detailed explanation

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,
\[
\begin{aligned}
\pi & =C R T=0.277 \times 0.0821 \times 288 \\
& =6.549 \simeq 6.57 \mathrm{~atm}
\end{aligned}
\]