# 0.4g of a gas has a volume of 227cm^3 at 27°C and at a pressure of 100kPa. Calculate the relative molecular mass of the gas (R=8.31 J/kmol)?

Sep 10, 2015

${M}_{r} = 43.95$

#### Explanation:

$P V = n R T$

$n = \frac{p V}{R T}$

$= \frac{100 \times {10}^{3} \times 227 \times {10}^{- 6}}{8.31 \times 300}$

$= 0.0091$

$n = \frac{m}{M} _ r$

${M}_{r} = \frac{m}{n} = \frac{0.4}{0.0091} = 43.95$

It looks like it could be $C {O}_{2}$ but there are other possibilities.

Sep 10, 2015

Molecular mass: $\text{44 g/mol}$.

#### Explanation:

To solve this problem you need to use the ideal gas law equation

$\textcolor{b l u e}{P V = n R T} \text{ }$, where

$P$ - the pressure of the gas;
$V$ - the volume of th gas;
$n$ - the number of moles of gas;
$R$ - the universal gas constant, in your case given as $\text{8.31 J/mol K}$
$T$ - the temperature of the gas - expressed in Kelvin!

The trick here is to realize that the units of $R$ can be rewritten as

R = "J"/("mol" * "K") = ("kg" * "m"^2)/"s"^2 * 1/("mol" * "K")

="kg"/("m" * "s"^2) * "m"^3 * 1/("mol" * "K")

$= \left(\text{Pa" * "m"^3)/("mol" * "K}\right)$

To find the molecular mass of the gas, you need to know how many moles of gas have a mass of 0.4 g. Rearrange the ideal gas law equation to solve for $n$

$n = \frac{P V}{R T}$

Next, convert the pressure from kPa to Pa and the volume from cubic centimeters to cubic meters

100color(red)(cancel(color(black)("kPa"))) * "1000 Pa"/(1color(red)(cancel(color(black)("kPa")))) = 10^5"Pa"

and

227color(red)(cancel(color(black)("cm"^3))) * ("1 m"""^3)/(10^6color(red)(cancel(color(black)("cm"^3)))) = 227 * 10^(-6)"m"^3

Plug in these values to find $n$, but do not forget to convert the temperature from degrees Celsiu to Kelvin!

n = (10^5color(red)(cancel(color(black)("Pa"))) * 227 * 10^(-6)color(red)(cancel(color(black)("m"^3))))/(8.31(color(red)(cancel(color(black)("Pa"))) * color(red)(cancel(color(black)("m"^3))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 27)color(red)(cancel(color(black)("K")))) = "0.00910 moles"

The molar mass of the gas will be

M_M = m/n = "0.4 g"/"0.00910 moles" = color(green)("44 g/mol")

SIDE NOTE I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of the gas.