Calculate the osmotic pressure of a 6.0 × 10-2 M solution of NaCl at 20°C (293 K).
Okay when I set up problem MRT....I get something totally different from the answer 2.9 atm. Could u please show me your set up to get to 2.9 atm? Thanks a billion!!
TutorCat gave iMRT. I'll bet you didn't include i which is 2 for NaCl. Multiply your answer by 2 and see if that doesn't do it.
Sure! To calculate the osmotic pressure using the van't Hoff equation (MRT equation), we can use the formula:
π = MRT
Where:
π is the osmotic pressure (in atm)
M is the molarity of the solution (in mol/L)
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin (K)
Given:
Molarity (M) = 6.0 × 10^-2 M
Temperature (T) = 20°C = 293 K
Let's substitute the given values into the formula:
π = MRT
= (6.0 × 10^-2 M) × (0.0821 L·atm/(mol·K)) × (293 K)
Now, let's calculate:
π = (6.0 × 10^-2) × (0.0821) × (293)
≈ 1.433 atm
It seems there was an error in the original calculation. The correct osmotic pressure for this solution is approximately 1.433 atm, not 2.9 atm.
To calculate the osmotic pressure (π) of a solution, you can use the formula π = MRT, where M is the molarity of the solution, R is the ideal gas constant (0.0821 L·atm/(K·mol)), and T is the temperature in Kelvin.
Given:
Molarity of NaCl solution (M) = 6.0 × 10^-2 M
Temperature (T) = 20°C = 293 K
To get the osmotic pressure:
1. Convert the temperature from Celsius to Kelvin:
T(K) = 20°C + 273 = 293 K
2. Plug in the values into the formula:
π = MRT
= (6.0 × 10^-2 M) × (0.0821 L·atm/(K·mol)) × (293 K)
3. Calculate the osmotic pressure:
π = (6.0 × 10^-2) × (0.0821) × (293)
≈ 14.229 atm
Therefore, the osmotic pressure of the 6.0 × 10^-2 M NaCl solution at 20°C (293 K) is approximately 14.229 atm.
If your result is significantly different from 2.9 atm, ensure that you performed the calculations correctly, including the proper conversion of temperature to Kelvin, and carefully recheck your calculations.