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

Answer: (A)

6.57 atm

Answer 2

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.