Question

Question #3ed00

Answer 1

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, `"H"_2`.

In your case, you have

`"VD" = "molar mass chloride"/"molar mass H"_2 = 59.5`

To make the calculations easier, you can use the approximation

`"molar mass H"_2 ~~ "2 g mol"^(-1)`

This means that you have

`"molar mass chloride" = 59.5 * "2 g mol"^(-1)`

`"molar mass chloride" = "119 g mol"^(-1)`

Now, you know that the equivalent mass of the metal is equal to `"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 `"35.5 g"` of chlorine, `"Cl"`.

So if `"24 g"` of the metal combine with `"35.5 g"` of chlorine, it follows that you have

`"24 g metal + 35.5 g Cl = 59.5 g chloride"`

`"24 g metal + 35.5 g Cl = 59.5 g chloride"`
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`"48 g metal + 71 g Cl = 119 g chloride"`

This means that `"48 g"` of metal combine with `"71 g"` of chlorine to form `"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

`"molar mass metal = 48 g mol"^(-1)`

The relative atomic mass of the metal will thus be

`A_r = (48 color(red)(cancel(color(black)("g mol"^(-1)))))/(1color(red)(cancel(color(black)("g mol"^(-1))))) = 48`