What is the molecular mass of a gas if 3.7 g of the gas at 25 °C occupy the same volume as 0.184 g of hydrogen at 17 °C and the same pressure?

Aug 15, 2017

The molecular mass of the gas is 42 u.

Explanation:

We can use the Ideal Gas Law to solve this problem:.

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Since $n = \frac{m}{M}$, we can rearrange this equation to get

$p V = \left(\frac{m}{M}\right) R T$

And we can solve this equation to get

$M = \frac{m R T}{p V}$

In this problem, we have two gases.

We can write

${M}_{1} = \frac{{m}_{1} R {T}_{1}}{p V}$ and ${M}_{2} = \frac{{m}_{2} R {T}_{2}}{p V}$

Dividing ${M}_{2}$ by ${M}_{1}$, we get

M_2/M_1 = (m_2color(red)(cancel(color(black)(R)))T_2)/(color(red)(cancel(color(black)(pV)))) × (color(red)(cancel(color(black)(pV))))/(m_1color(red)(cancel(color(black)(R)))T_1) = (m_2T_2)/(m_1T_1)

M_2 = M_1 × m_2/m_1 × T_2/T_1

In this problem,

M_1 = "2.016 u";color(white)(mmmmmmmmml) M_2 = ?
${m}_{1} = \text{0.184 g";color(white)(mmmmmmmmmll) m_2 = "3.7 g}$
${T}_{1} = \text{(17 + 273.15) K" = "290.15 K"; T_2 = "(25 + 273.15) K" = "298.15 K}$

${M}_{2} = \text{2.016 u" × (3.7 color(red)(cancel(color(black)("g"))))/(0.184 color(red)(cancel(color(black)("g")))) × (298.15 color(red)(cancel(color(black)("K"))))/(290.15 color(red)(cancel(color(black)("K")))) = "42 u}$