# Consider a solution containing 4.62 g of a Group I metal carbonate, M2CO3 in a total volume of 100.0 mL. If 25.0 mL of this solution requires 20.0 mL of 0.5 mol L1 HCl for complete reaction, the Relative Atomic Mass of the metal M is?

##### 1 Answer

See below:

#### Explanation:

In a reaction of carbonates and acids we have the general formula:

Carbonate + Acid -> Salt+ Carbon dioxide + water

In this case:

I'm going to assume that if we take 25ml of the 100 ml solution, we will get

Lets find our how many moles of

Hence

And from the reaction equation we see that this is twice the amount of moles of carbonate reacted- so there was only

Now we can find the molar mass of the carbonate, as we know that we used

Hence:

This is the molar mass of the carbonate, so now we just need to subtract the molar mass of the known elements from it. We have three oxygen atoms and one carbon atom, so we subtract the molar masses of these elements from the total:

Consequently:

Which seems to lie very close to element 37, Rubidium (

So the carbonate must have been Rubidium carbonate: