Question

If 3 g of glucose (Molar mass = 180) is dissolved in 60 g of water at 15 degree, the osmotic pressure of the solution will be

• A. 6.57 atm
• B. 0.34 atm
• C. 5.57 atm
• D. 0.65 atm

Answer 1

The Correct Option is A

To calculate the osmotic pressure of a solution, we can use the formula:
\(\text{Osmotic pressure} = \left( \frac{n}{V} \right)RT\)
First, let's calculate the number of moles of glucose:
\(\text{Number of moles of glucose} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}}\)

\(\text{Number of moles of glucose} = \frac{3 \, \text{g}}{180 \, \text{g/mol}} = 0.0167 \, \text{mol}\)

Next, we need to convert the mass of water to volume:
\(\text{Volume of water} = \frac{\text{mass of water}}{\text{density of water}}\)

\(\text{Volume of water} = \frac{60 \, \text{g}}{1 \, \text{g/cm}^3} = 60 \, \text{cm}^3 = 0.06 \, \text{L}\)

Now, we can calculate the osmotic pressure:
\(\text{Osmotic pressure} = \left( \frac{0.0167 \, \text{mol}}{0.06 \, \text{L}} \right) \times (0.0821 \, \text{L.atm/(mol.K)}) \times (15 + 273) \, \text{K}\)

\(\text{Osmotic pressure} \approx 6.57 \, \text{atm}\)
Therefore, the osmotic pressure of the solution will be approximately (A) 6.57 atm.

Answer 2

Given: 

• Mass of glucose = 3 g
• Molar mass of glucose = 180 g/mol
• Mass of water = 60 g = 0.060 kg
• Temperature = 15°C = 288 K
• R = 0.0821 L·atm/mol·K

First, calculate moles of glucose:
$$ \text{Moles} = \frac{3}{180} = 0.01667 \, \text{mol} $$

Volume of solvent ≈ volume of solution (since water is solvent)
Assume density ≈ 1 g/mL ⇒ 60 g = 60 mL = 0.060 L

Now apply the formula for osmotic pressure:
$$ \Pi = \frac{n}{V}RT $$ where \( n = 0.01667 \), \( V = 0.060 \, \text{L} \), \( R = 0.0821 \), \( T = 288 \, \text{K} \)

Substitute values:
$$ \Pi = \frac{0.01667}{0.060} \cdot 0.0821 \cdot 288 $$

$$ \Pi ≈ 0.2778 \cdot 0.0821 \cdot 288 ≈ 0.2778 \cdot 23.6448 ≈ 6.57 \, \text{atm} $$

Correct answer: 6.57 atm