What is the molar mass of a gas if 3.50 grams of the gas occupy 1.35 liters of space at STP?

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What is the molar mass of a gas if 3.50 grams of the gas occupy 1.35 liters of space at STP?

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Answer 1 · 2018-06-18T01:39:46

The molar mass of the gas is $58.9 g/mol$ $"58.9 g/mol"$ .

Explanation:

You will need to use the equation for the ideal gas law:

$P V = n R T$ $PV=nRT$ ,

where:

$P$ $P$ is pressure, $V$ $V$ is volume, $n$ $n$ is moles, $R$ $R$ is gas constant, and $T$ $T$ is temperature.

$STP$ $"STP"$ is $0^{\circ} C$ $0^@"C"$ or $273.15 K$ $"273.15 K"$ (required for gas laws), and $10^{5}$ $10^5$ $Pa$ $"Pa"$ or $100 kPa$ $"100 kPa"$ .

Use the equation for the ideal gas law to calculate moles of gas. Then calculate the molar mass by dividing the given mass by the calculated moles.

Known

$P = 100 kPa$ $P="100 kPa"$

$V = 1.35 L$ $V="1.35 L"$

$R = {8.31446 L kPa K}^{- 1} {mol}^{- 1}$ $R="8.31446 L kPa K"^(-1) "mol"^(-1)"$

$T = 273.15 K$ $T="273.15 K"$

Unknown

$n$ $n$

Solution

Rearrange the ideal gas law equation to isolate $n$ $n$ . Plug in the known values and solve.

$n = \frac{P V}{R T}$ $n=(PV)/(RT)$

$n=(100color(red)(cancel(color(black)("kPa")))xx1.35color(red)(cancel(color(black)("L"))))/(8.31446 color(red)(cancel(color(black)("L"))) color(red)(cancel(color(black)("kPa"))) color(red)(cancel(color(black)("K")))^(-1) "mol"^(-1)xx273.15color(red)(cancel(color(black)("K"))))="0.0594 mol"$ (rounded to three significant figures)

To calculate the molar mass of the gas, divide its given mass in grams by the calculated number of moles.

$"molar mass"=("3.50 g")/("0.0594 mol")="58.9 g/mol"$

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What is the molar mass of a gas if 3.50 grams of the gas occupy 1.35 liters of space at STP?

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