Question

A 3.0 g sample of a gas occupies a volume of 1.00L at 100C and 740 torr pressure. The molecular weight of the gas is ______________________. (Give the same ANSWER for the next TWO problems) 8. The molecular weight is a . 125 g/mol d . 94.3 g/mol b . 25.3 g/mol e . 54.0 g/mol c . 74.2 g/mol

Answer 1

Answer:

94.3g/mol

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Mass of the gas = 3g

Volume (V) = 1L

Temperature (T) = 100°C

Pressure (P) = 740torr

Molecular weight of the gas =..?

Step 2:

Conversion to appropriate unit.

For temperature:

T(K) = T(°C) + 273

T(°C) = 100°C

T(K) = 100°C+ 273

T(K) = 373K

For Pressure:

760torr = 1atm

Therefore, 740torr = 740/760 = 0.974 atm

Step 3:

Determination of the number of mole of the gas.

The number of mole of the gas can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

Volume (V) = 1L

Temperature (T) = 373K

Pressure (P) = 0.974atm

Number of mole (n) =..?

Gas constant (R) = 0.0821atm.L/Kmol

PV = nRT

Divide both side by RT

n = PV / RT

n = 0.974 x 1 / 0.0821 x 373

n = 0.0318 mole

Step 4:

Determination of the molecular weight of the gas. This can be obtained as shown below:

Mass of gas = 3g

Number of mole of gas = 0.0318 mol

Molecular weight =?

Molecular weight = Mass /number of mole

Molecular weight of the gas = 3/0.0318

Molecular weight of the gas = 94.3g/mol