Determine the atomic weight of X in the following compounds
a)8g XH4 occupy 44.8L volume at 237C and 380 mmHg pressure
b)1.92g of X3 occupis a volume of 574 ml at 77C and 2 atm
c)A O.326 gram of gas,C3X4 has a volume of 250 ml at100C and 76 cmHg
d)A 3.12 gram of the gaS.HX,occupies a volume of 500 ml at -23C and 760 torr
Use PV = nRT and solve for n = number of moles. Then n = grams/molar mass, solve for molar mass.
To determine the atomic weight of X in the given compounds, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles
R = Gas constant
T = Temperature
Let's go through each compound and calculate the atomic weight of X. We'll start with compound a.
a) 8g XH4 occupying 44.8L volume at 237°C and 380 mmHg pressure:
First, we need to convert the given pressure from mmHg to atm. We divide 380 mmHg by 760 mmHg/atm to get 0.5 atm.
Using the ideal gas law equation, rearranged to isolate n:
n = (PV) / (RT)
Substituting the given values:
n = (0.5 atm * 44.8 L) / (0.0821 atm L/mol K * 510 K)
n ≈ 0.535 mol
We know that XH4 contains one mole of X, so the atomic weight of X is equal to the molar mass of XH4, divided by the number of moles:
Atomic weight of X = (8g) / (0.535 mol)
Atomic weight of X ≈ 14.95 g/mol
b) 1.92g of X3 occupying a volume of 574 ml at 77°C and 2 atm:
Again, let's convert the given pressure from atm to mmHg. Multiply 2 atm by 760 mmHg/atm to get 1520 mmHg.
Applying the ideal gas law equation:
n = (PV) / (RT)
Substituting the values:
n = (1520 mmHg * 0.574 L) / (0.0821 atm L/mol K * 350 K)
n ≈ 32.05 mol
Since X3 contains three moles of X, the atomic weight of X is calculated as:
Atomic weight of X = (1.92g) / (32.05 mol/3)
Atomic weight of X ≈ 0.179 g/mol
c) 0.326 gram of gas, C3X4, with a volume of 250 ml at 100°C and 76 cmHg:
Converting the pressure from cmHg to atm, we divide 76 cmHg by 760 cmHg/atm, resulting in 0.1 atm.
Using the ideal gas law equation:
n = (PV) / (RT)
Substituting the values:
n = (0.1 atm * 0.250 L) / (0.0821 atm L/mol K * 373 K)
n ≈ 0.0085 mol
As C3X4 contains three moles of X, the atomic weight of X becomes:
Atomic weight of X = (0.326g) / (0.0085 mol/3)
Atomic weight of X ≈ 38.47 g/mol
d) 3.12 gram of the gas HX, occupying a volume of 500 ml at -23°C and 760 torr:
Converting the temperature from Celsius to Kelvin, we add 273 to -23 to get 250K. Also, we convert the pressure from torr to atm by dividing 760 torr by 760 torr/atm to obtain 1 atm.
Applying the ideal gas law equation:
n = (PV) / (RT)
Substituting the values:
n = (1 atm * 0.500 L) / (0.0821 atm L/mol K * 250 K)
n ≈ 0.0242 mol
The atomic weight of X is equivalent to the molar mass of HX divided by the number of moles:
Atomic weight of X = (3.12g) / (0.0242 mol)
Atomic weight of X ≈ 128.51 g/mol
Therefore, the atomic weight of X in the given compounds is approximately as follows:
a) 14.95 g/mol
b) 0.179 g/mol
c) 38.47 g/mol
d) 128.51 g/mol