If a 2.15-g sample of gas occupies 750. mL at STP, what is the molar mass of the gas at 125 degrees celcuis
We cannot determine the molar mass of the gas at 125 degrees Celsius directly, but we can determine the molar mass of the gas at STP and then apply that value to the given conditions at 125 degrees Celsius. Here's how:
At STP (Standard Temperature and Pressure), the temperature is 0 degrees Celsius (273.15 K) and the pressure is 1 atm.
1. Determine the number of moles of the gas using the ideal gas law:
PV = nRT
We know the following at STP:
P = 1 atm
V = 750 mL = 0.750 L (conversion)
R = 0.0821 L*atm/mol*K (ideal gas constant)
T = 273.15 K
Plug in the values to find the number of moles (n):
(1 atm) * (0.750 L) = n * (0.0821 L*atm/mol*K) * (273.15 K)
Solve for n:
n = (1 atm * 0.750 L) / (0.0821 L*atm/mol*K * 273.15 K)
n ≈ 0.03271 moles
2. Calculate the molar mass (MM) of the gas:
MM = mass of the sample / number of moles
MM = 2.15 g / 0.03271 moles
MM ≈ 65.75 g/mol
Now we have the molar mass of the gas (65.75 g/mol) at STP, assuming that the molar mass does not change under different conditions, we can report that under 125 degrees Celsius, the molar mass of the gas is also approximately 65.75 g/mol.
To find the molar mass of the gas at 125 degrees Celsius, we need to follow a few steps:
Step 1: Convert the given mass of the gas from grams to moles.
- Given mass of the gas = 2.15 g
Using the molar mass equation: moles = mass / molar mass
- moles = 2.15 g / molar mass
Step 2: Convert the given volume of the gas from milliliters to liters.
- Given volume of the gas = 750 mL
Using the conversion factor: 1 L = 1000 mL
- volume = 750 mL * (1 L / 1000 mL) = 0.750 L
Step 3: Apply the ideal gas law equation to find the molar mass.
- The ideal gas law equation: PV = nRT
At STP (standard temperature and pressure), the values are:
- P = 1 atm
- V = 22.4 L (molar volume of a gas at STP)
- T = 273 K
Looking for the value of n (moles), we can rearrange the ideal gas law equation:
n = (PV) / (RT)
n = (1 atm * 0.750 L) / (0.0821 L·atm/(mol·K) * (273 + 125) K)
n = 0.0099 mol
Step 4: Solve for the molar mass.
- Apply the equation from Step 1 to find the molar mass:
molar mass = mass / moles
molar mass = 2.15 g / 0.0099 mol
molar mass = 217.17 g/mol
Therefore, the molar mass of the gas at 125 degrees Celsius is approximately 217.17 g/mol.
To find the molar mass of the gas, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. At STP (standard temperature and pressure), the pressure is 1 atm, and the temperature is 0 degrees Celsius or 273 kelvin.
First, we need to find the number of moles of gas at STP. We can use the equation n = PV/RT, where P = 1 atm, V = 750 mL (which we need to convert to liters, so 750 mL = 0.750 L), R = 0.0821 L·atm/mol·K (ideal gas constant), and T = 273 K.
n = (1 atm * 0.750 L) / (0.0821 L·atm/mol·K * 273 K)
n ≈ 0.031 mol
Next, we need to find the molar mass. The formula for molar mass is M = m/n, where M is the molar mass, m is the mass of the gas, and n is the number of moles.
Given that the mass of the gas is 2.15 g and n ≈ 0.031 mol, we can calculate the molar mass of the gas.
M = (2.15 g) / (0.031 mol)
M ≈ 69.3 g/mol
Therefore, the molar mass of the gas at 125 degrees Celsius is approximately 69.3 g/mol.