Question

Determine the mass (in g) of oxygen (mr = 32.00) that occupies a volume of 146.7 cm3 at a pressure of 106.5 kpa and a temperature of 167 oc.

Answer 1

Answer:- 0.1366 g

Solution:- It's based on an ideal gas law equation, Using ideal gas law equation the moles of gas are calculated and on multiplying by molar mass we get the grams of the gas.

Given, V = $146.7cm^3$

P = 106.5 kpa

T = 167+273 = 440 K

Pressure unit also needs to be converted by kpa to atm.

1 atm = 101.325 kpa

So, $106.5kpa(\frac{1atm}{101.325kpa})$

= 1.051 atm

Volume needs to be converted to L and for this we divide by 1000 since 1 L equals to 1000 cubic centimeter.

$146.7cm^3(\frac{1L}{1000cm^3})$

= 0.1467 L

An ideal gas law equation is, PV = nRT

$n=\frac{PV}{RT}$

where R is the universal gas law constant and its value is 0.0821 atm.L per mol per kelvin.

Let's plug in the values in the equation:

$n=\frac{1.051atm*0.1467L}{\frac{0.0821atm.L}{mol.K}*440K}$

n = 0.004268 mol

let's multiply the moles by molar mass to get the grams of the oxygen gas.

$0.004268mol(\frac{32.00g}{mol})$

= 0.1366 g

So, the mass of the oxygen gas is 0.1366 g.