calculate the osmotic pressure of 5% glucose solution,where the molecular weight of glucose is 180 gm,at 18 c ??
The Correct Answer and Explanation is :
To calculate the osmotic pressure of a 5% glucose solution at 18°C, we will use the formula for osmotic pressure:
[
\Pi = \frac{nRT}{V}
]
where:
- (\Pi) is the osmotic pressure in atmospheres (atm),
- (n) is the number of moles of solute,
- (R) is the ideal gas constant (0.0821 L·atm/mol·K),
- (T) is the absolute temperature in Kelvin (K),
- (V) is the volume of the solution in liters.
Step-by-Step Calculation:
- Convert the mass percentage of glucose to moles: A 5% glucose solution means there are 5 grams of glucose per 100 mL of solution. Since we are dealing with 100 mL, it’s equivalent to 0.1 L. To calculate the number of moles of glucose: [
\text{Moles of glucose} = \frac{\text{Mass of glucose}}{\text{Molecular weight of glucose}}
] Given:
- Mass of glucose = 5 grams,
- Molecular weight of glucose = 180 g/mol. [
\text{Moles of glucose} = \frac{5 \text{ g}}{180 \text{ g/mol}} = 0.02778 \text{ moles}
]
- Convert the temperature to Kelvin: The temperature is given as 18°C. To convert this to Kelvin: [
T = 18°C + 273.15 = 291.15 \text{ K}
] - Calculate the osmotic pressure: Now, we can substitute all the values into the osmotic pressure formula: [
\Pi = \frac{(0.02778 \text{ mol})(0.0821 \text{ L·atm/mol·K})(291.15 \text{ K})}{0.1 \text{ L}}
] [
\Pi = \frac{(0.02778)(0.0821)(291.15)}{0.1}
] [
\Pi \approx 0.0673 \text{ atm}
]
Explanation:
The osmotic pressure of a solution is a colligative property, meaning it depends on the number of solute particles in the solution, not the identity of the solute. The formula (\Pi = \frac{nRT}{V}) tells us how much pressure is exerted by the solute particles when they move across a semipermeable membrane. The greater the number of solute particles, the higher the osmotic pressure.
In this case, we calculated the osmotic pressure of a 5% glucose solution at 18°C and found it to be approximately 0.0673 atm.