# Question b629a

May 22, 2017

${\text{4.3M g mol}}^{- 1}$

#### Explanation:

For starters, it's worth mentioning that NTP conditions are defined as a pressure of $\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 atm}}}}$ and a temperature of

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{20}^{\circ} \text{C"))) = 20^@"C" + 273.15 = color(blue)(ul(color(black)("293.15 K}}}}$

Now, use the ideal gas law equation to calculate the molar volume of the gas at NTP, i.e. the volume occupied by $1$ mole of any ideal gas under NTP conditions.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Rearrange the equation

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

Plug in your values to find

V/n = (0.082 (color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * 293.15 color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm"))))

$\frac{V}{n} = {\text{24.04 L mol}}^{- 1}$

This tells you that $1$ mole of any ideal gas occupies $\text{24.04 L}$ under NTP conditions.

You can thus say that your sample, which is said to occupy $\text{5.6 L}$ under these conditions, will contain

5.6 color(red)(cancel(color(black)("L"))) * "1 mole"/(24.04color(red)(cancel(color(black)("L")))) = "0.2329 moles"

As you know, the molar mass of a substance is defined as the mass of exactly $1$ mole of said substance.

In your case, $1$ mole of this substance will have a mass of

1 color(red)(cancel(color(black)("mole"))) * "M g"/(0.2329 color(red)(cancel(color(black)("moles")))) = (1/0.2329 * "M")# $\text{g}$

Therefore, you can say that the molar mass of the substance is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molar mass" = (4.3 * "M")color(white)(.)"g mol}}^{- 1}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the sample.