A 5.0 g n2o (laughing gas) that is inside a 12 mL cylinder used used to fill up a 800 mL whipped cream container. Beforehand, the container had air in it at 760 torr atmosphere pressure at 25 degrees celsius. Then, we will add the n2o/air mix to balloon until the pressure in the whipped cream container and the filled balloon are both at 760 torr. a) What is the mole fraction of n2o in the whipped cream container?
b) estimate the decimal fraction of the 5.0 g n2o that ends up in the balloon.
Thanks in advance for help!
Is all 5.0 g N2O transferred to to the whipped cream container? And what is the volume of the balloon?
To calculate the mole fraction of N2O in the whipped cream container, you first need to determine the moles of N2O and air separately.
a) Mole fraction of N2O in the whipped cream container:
1. Convert the mass of N2O from grams to moles using its molar mass.
The molar mass of N2O is 44.013 g/mol.
Moles of N2O = 5.0 g / 44.013 g/mol = 0.1136 mol
2. To calculate the moles of air, you need to know the total moles of gas in the container at the beginning. We can use the ideal gas law, assuming ideal behavior:
PV = nRT
P: Pressure (760 torr)
V: Volume (12 mL)
n: Moles of gas (unknown)
R: Ideal gas constant (0.0821 L·atm/(mol·K))
T: Temperature (25 °C = 298 K)
Rearrange the equation to solve for n:
n = (PV) / (RT)
n = (760 torr * 12 mL) / (0.0821 L·atm/(mol·K) * 298 K)
Remember to convert mL to L by dividing by 1000.
n = 0.3628 mol
3. Now that you know the moles of N2O and air, sum them to get the total moles of gas in the container:
Total moles = Moles of N2O + Moles of air
Total moles = 0.1136 mol + 0.3628 mol
Total moles ≈ 0.4764 mol
4. Finally, calculate the mole fraction of N2O:
Mole fraction of N2O = Moles of N2O / Total moles
Mole fraction of N2O = 0.1136 mol / 0.4764 mol ≈ 0.2385
Therefore, the mole fraction of N2O in the whipped cream container is approximately 0.2385.
b) To estimate the decimal fraction of the 5.0 g N2O that ends up in the balloon, we need to know the molar volume of the gases involved. At STP (Standard Temperature and Pressure: 0 °C and 1 atm), 1 mole of any ideal gas occupies 22.414 L.
1. Estimate the initial volume of N2O in the cylinder using its molar volume at STP.
Moles of N2O = 0.1136 mol
Volume of N2O at STP = Moles of N2O * Molar volume at STP
= 0.1136 mol * 22.414 L
≈ 2.542 L
2. Determine the initial total volume of the gas mixture in the whipped cream container.
Initial total volume = Volume of N2O + Volume of air
Initial total volume = 2.542 L + 0.012 L (converted 12 mL of container volume to L)
Initial total volume ≈ 2.554 L
3. Since the gas is transferred to a balloon until the pressure reaches 760 torr, we can assume the volume remains constant during the transfer.
4. Calculate the mole fraction of N2O in the balloon. It will be the same as the mole fraction in the whipped cream container.
Mole fraction of N2O = 0.2385
5. Estimate the moles of N2O in the balloon.
Moles of N2O in the balloon = Mole fraction of N2O * Total moles in the balloon
To calculate the total moles in the balloon, use the ideal gas law as done in the calculation for part (a).
6. Finally, estimate the decimal fraction of the 5.0 g N2O present in the balloon by dividing the moles of N2O in the balloon by the moles of N2O initially present:
Decimal fraction of N2O in the balloon = (moles of N2O in the balloon) / (moles of N2O initially present)
Although specific calculations were not provided for part (b), following these steps will help you estimate the decimal fraction of the 5.0 g N2O that ends up in the balloon.