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A sample of gas of mass 2.929g occupies a volume of 426mL at 0°C and 1.00atm pressure.what is molecular weight of the gas?

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Answer:

154 g/mole

Step-by-step explanation:

We are given the mass of the gas, but we also need the number of moles the 2.929g represents. Since we are provided the conditions of the gas, we can the Ideal Gas law to find the number of moles of the mystery gas.

Ideal Gas Law: PV = nRT, where P, V, and T are the pressure, volume, and temperature (temperature must be in degrees Kelvin), n is the moles, and R is the gas constant.

Let's choose the gas constant that has the same units as we were given. R = 0.0820575 [L⋅atm⋅/(K⋅mol)] comes the closest, but we'll still need to convert ml to liters(L) and °C to °K:

426mL = 0.426L

0°C = 273.25 [add 273.15 to the Centrigrade value]

Let's rearrnage the ideal gas law to solve for n, the number of moles:

n = (PV/RT)

Now enter the data:

n = (1atm)(0.426L)/[(0.0820575 L⋅atm⋅/(K⋅mol))*(273.15°K)]

n = (1atm)(0.426L)/[(0.0820575 Latm/(K⋅mol))*(273.15°K)] [Units that cancle are highlighted]

n = (1)(0.426)/[(0.0820575 /(mol))*(273.15)] [We are left only with moles (mol)

n = (0.426)/(0.0820575)/(273.15) [1/1/mol] [Move the only unit out (1/1/mol)]

n = (0.426)/(0.0820575)/(273.15) [1/1/mol] = 0.0190 moles

Note that the unit moves to the top, i.e., : 1/1/mol = mole

We have the mass and the number of moles. Divide the two to obtain molar mass:

(2.929g)/(0.0190 moles) = 154 g/mole This is also the molecular weight.

[I don't know what is a gas at 0°C and has that molecular weight]

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