A mixture of nitrogen gas and water vapour axis contained in a gas cylinder and the pressure of the gas mixture was found to be 942mmHg, what is the pressure exerted from the water vapour if the partial pressure exerted by the nitrogen gas is 797mmHg.
Ptotal = pN2 + pH2O
You now Ptotal and pN2, solve for pH2O
What's water vapor "axis"?
A gaseous sample weighing 0.896g was found to occupy a volume of 0.524dm3 at 0.96atm and 28°C
1. What is the molar mass of the gas
2. What gas is it
Use PV = nRT. You know P, V, R (use 0.08206 dm^3*atm/mol*K) and T (remember to convert degrees C to kelvin). Solve for n = number of mols.
Then n = grams/molar mass. Knowing n and grams, calculate molar mass).
I'll leave it to you to figure out the gas.
To determine the pressure exerted by the water vapor, we can subtract the partial pressure of nitrogen gas from the total pressure of the gas mixture.
The pressure exerted by the water vapor can be calculated using the formula:
Pressure of water vapor = Total pressure - Partial pressure of nitrogen gas
Given that the total pressure is 942 mmHg and the partial pressure of nitrogen gas is 797 mmHg, we can substitute these values into the formula:
Pressure of water vapor = 942 mmHg - 797 mmHg
Pressure of water vapor = 145 mmHg
Therefore, the pressure exerted by the water vapor in the gas mixture is 145 mmHg.