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 L1 HCl for complete reaction, the Relative Atomic Mass of the metal M is?

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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 L1 HCl for complete reaction, the Relative Atomic Mass of the metal M is?

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Answer 1 · 2018-04-22T17:42:23

See below:

Explanation:

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

Carbonate + Acid -> Salt+ Carbon dioxide + water

In this case:

$M_{2} C O_{3} (a q) + 2 H C l (a q) \to 2 M C l (a q) + C O_{2} (g) + H_{2} O (l)$ $M_2CO_3(aq)+2HCl(aq) -> 2MCl(aq)+CO_2(g)+H_2O(l)$

I'm going to assume that if we take 25ml of the 100 ml solution, we will get $\frac{1}{4}$ $1/4$ of the mass of carbonate dissolved in the solution- 1.155 g.

Lets find our how many moles of $H C l$ $HCl$ reacted with the carbonate using the concentration equation:

$c = \frac{n}{v}$ $c=(n)/v$

$c$ $c$ = 0.5 $m o l d m^{- 3}$ $moldm^-3$
$n$ $n$ =unknown
$v$ $v$ =20 ml =0.02 liters / $d m^{3}$ $dm^3$

Hence
$n = c v$ $n=cv$

$n = 0.02 \times 0.5$ $n=0.02 times 0.5$

$n = 0.01$ $n=0.01$

And from the reaction equation we see that this is twice the amount of moles of carbonate reacted- so there was only $0.005$ $0.005$ mol of carbonate that reacted in the reaction.

Now we can find the molar mass of the carbonate, as we know that we used $0.005$ $0.005$ mol and it had a mass of 1.155 grams.

$M = \frac{m}{n}$ $M=(m)/n$

$M$ $M$ =molar mass
$m$ $m$ =mass in grams
$n$ $n$ =number of moles

Hence:

$M = \frac{1.155}{0.005}$ $M=(1.155)/0.005$

$M = 231 g m o l^{- 1}$ $M=231 gmol^-1$

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:

$231 - 3 (16) - 12 = 171 g m o l^{- 1}$ $231-3(16)-12=171 gmol^-1$

Consequently:

$2 M = 171 g m o l^{- 1}$ $2M=171 gmol^-1$

$M = 85.5 g m o l^{- 1}$ $M=85.5 gmol^-1$

Which seems to lie very close to element 37, Rubidium ( $R b$ $Rb$ ) with a molar mass of $85.47 g m o l^{- 1}$ $85.47 gmol^-1$ , which is also a group 1 metal.
So the carbonate must have been Rubidium carbonate:

$R b_{2} C O_{3}$ $Rb_2CO_3$

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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 L1 HCl for complete reaction, the Relative Atomic Mass of the metal M is?

1 Answer

Explanation:

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