A 0.88\text{ g} sample of a purified gas in a laboratory experiment occupies a volume of 4… — JAMB UTME

A $0.88\text{ g}$ sample of a purified gas in a laboratory experiment occupies a volume of $400\text{ cm}^3$ at $27^\circ\text{C}$ and a pressure of $1.23\text{ atm}$ . Given that the gas constant $R = 0.082\text{ dm}^3\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ , what is the molar mass of the gas?

$44\text{ g mol}^{-1}$ Answer
B
$4.0\text{ g mol}^{-1}$
C
$49.3\text{ g mol}^{-1}$
D
$0.044\text{ g mol}^{-1}$

Answer

The molar mass of the gas is

44\text{ g mol}^{-1}

Converting all parameters to standard SI/gas units yields

V = 0.40\text{ dm}^3

and

T = 300\text{ K}

. Substituting these into the ideal gas equation

n = \frac{PV}{RT}

gives

n = 0.020\text{ mol}

. Dividing the sample mass (

0.88\text{ g}

) by the calculated number of moles gives the correct molar mass of

44\text{ g mol}^{-1}

Step-by-Step Solution

Convert given values into appropriate units matching the gas constant

R

V = \frac{400}{1000} = 0.40\text{ dm}^3

T = 27 + 273 = 300\text{ K}

P = 1.23\text{ atm}

The ideal gas law requires temperature in Kelvin and volume in

\text{dm}^3

to match the units of

R

Calculate the number of moles (

n

) using the ideal gas equation

PV = nRT

n = \frac{PV}{RT} = \frac{1.23 \times 0.40}{0.082 \times 300} = \frac{0.492}{24.6} = 0.020\text{ mol}

Rearranging

PV = nRT

allows direct computation of gas moles.

Compute the molar mass (

M

) using

M = \frac{\text{mass}}{n}

M = \frac{0.88\text{ g}}{0.020\text{ mol}} = 44\text{ g mol}^{-1}

Molar mass is the ratio of given mass in grams to the amount of substance in moles.

Key Concept

Ideal Gas Equation and Molar Mass Determination

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