Question

QUESTION 10 What is the mole fraction of hydrogen in a gaseous mixture that consists of 8.00 g of hydrogen and 12.00 g of neon in a 3.50 liter container maintained at 35.20 °C? O 0.150 O 0.400 O 0.660 O 0.870 O 0.930

Answer 1

Given, Mass of hydrogen = 8 g

Mass of neon = 12 g

Volume of the container, V = 3.50 L

Temperature, T = 35.20⁰ C = 35.20 + 273 = 308.2 K

Mole fraction is calculated using the formula, $\chi_{i} = P_{i}/P_{total}$

where $\chi_{i}$ is the mole fraction of the gas i , $P_{i}$ is the partial pressure of gas i , $P_{total}$ is the total pressure of the mixture of gases.

The total pressure of a mixture of gases is obtained from the sum of the partial pressures of each individual gas.i.e, here total pressure of this mixture is equal to sum of the partial pressures of hydrogen and neon gases.

$P_{total} = P_{hyd} + P_{neon}$

We can use the ideal gas law to calculate the partial pressure of each gas.i.e,

$PV = nRT$

$P_{hyd} = (n_{hyd}RT)/V$

Similarly, $P_{neon} = (n_{neon}RT)/V$

$n_{hyd}, n_{neon}$ are the number of moles of hydrogen and neon gases.

Number of moles = given mass of the gas / molar mass of the gas

Molar mass of hydrogen = 1 g/mol

Molar mass of neon = 20 g/mol

Number of moles of hydrogen gas, $n_{hyd}$ = (8 g) / (1 g/mol) = 8 mol

Number of moles of neon gas, $n_{neon}$ = (12 g) / (20 g/mol) = 0.6 mol

So, $P_{hyd} = (n_{hyd}RT)/V$

R = 0.0821 atm L mol^-1 K^-1

Substituting the values, $P_{hyd}$ = [ (8 mol) * (0.0821 atm L mol^-1 K^-1) * (308.2 K) ] / (3.50 L)

= 57.8359 atm

$P_{neon} = (n_{neon}RT)/V$

$P_{neon}$ = [ (0.6 mol) * (0.0821 atm L mol^-1 K^-1) * (308.2 K) ] / (3.50 L)

= 4.3376 atm

Now, $P_{total} = P_{hyd} + P_{neon}$

$P_{total}$ = 57.8359 atm + 4.3376 atm = 62.1735 atm

We have to find the mole fraction of hydrogen gas.

$\chi _{hyd}= P_{hyd}/P_{total}$

Substituting the values,  $\chi _{hyd}$ = (57.8359 atm) / (62.1735 atm) = 0.9302

Hence, mole fraction of hydrogen = 0.930

Last option is the correct one.