# Question #3ed00

Aug 8, 2017

The relative atomic mass of the element is $48$.

#### Explanation:

The idea here is that the vapor density of the chloride will help you determine its molar mass.

As you know, the vapor density of a gas is defined as the ratio that exists between the molar mass of the gas and the molar mass of hydrogen gas, ${\text{H}}_{2}$.

${\text{VD" = "molar mass chloride"/"molar mass H}}_{2} = 59.5$

To make the calculations easier, you can use the approximation

${\text{molar mass H"_2 ~~ "2 g mol}}^{- 1}$

This means that you have

${\text{molar mass chloride" = 59.5 * "2 g mol}}^{- 1}$

${\text{molar mass chloride" = "119 g mol}}^{- 1}$

Now, you know that the equivalent mass of the metal is equal to $\text{24 g}$. In your case, the equivalent mass of the metal, which is essentially the mass of $1$ equivalent of metal, tells you the mass of metal that combines directly with $\text{35.5 g}$ of chlorine, $\text{Cl}$.

So if $\text{24 g}$ of the metal combine with $\text{35.5 g}$ of chlorine, it follows that you have

$\text{24 g metal + 35.5 g Cl = 59.5 g chloride}$

$\text{24 g metal + 35.5 g Cl = 59.5 g chloride}$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
$\text{48 g metal + 71 g Cl = 119 g chloride}$

This means that $\text{48 g}$ of metal combine with $\text{71 g}$ of chlorine to form $\text{119 g}$ of metal chloride, which tells you that the mass of $1$ mole of metal, which is equal to $2$ equivalents of metal, is equal to

${\text{molar mass metal = 48 g mol}}^{- 1}$

The relative atomic mass of the metal will thus be

${A}_{r} = \left(48 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g mol"^(-1)))))/(1color(red)(cancel(color(black)("g mol}}^{- 1}}}}\right) = 48$