Stoichiometry: Comparing Ion Numbers in Different Sample Solutions

[WMDhamnekar]

Which sample contains the greater number of ions?

1) $100cm^3$ of 0.05 mol $dm^{-3} CuS$(cupric sulfide)

2)$100 cm^3$ of 0.20 mol $dm^{-3} MgSO_4$ magnesium sulfate

3)$50 cm^3$ of 0.40 mol $dm^{-3} RbCl$ rubidium chloride

4)$100 cm^3$ of 0.20 mol $dm^{-3} CaCl_2$ calcium chlorideHow to answer this question?

 

[I like Serena]

You are given a volume $V$ and a molar concentration $c=\frac n V$, where $n$ is the number of moles.

The number of molecules is $n=c\cdot V$ in moles.

For 1) we have $n=c\cdot V= 0.05\, \frac{\text{mol}}{\text{dm}^3}\cdot 100\,\text{cm}^3 = 0.05\, \frac{\text{mol}}{1000\,\text{cm}^3}\cdot 100\,\text{cm}^3 = 0.005\,\text{mol} \,\ce{Cu S}$ molecules.

We get $\ce{Cu^2+}$ and $\ce{S^2-}$ ions, so the number of ions is twice the number of molecules, which is $0.01\,\text{mol}$ ions.

 

[WMDhamnekar]

You are given a volume $V$ and a molar concentration $c=\frac n V$, where $n$ is the number of moles.

The number of molecules is $n=c\cdot V$ in moles.

For 1) we have $n=c\cdot V= 0.05\, \frac{\text{mol}}{\text{dm}^3}\cdot 100\,\text{cm}^3 = 0.05\, \frac{\text{mol}}{1000\,\text{cm}^3}\cdot 100\,\text{cm}^3 = 0.005\,\text{mol} \,\ce{Cu S}$ molecules.

We get $\ce{Cu^2+}$ and $\ce{S^2-}$ ions, so the number of ions is twice the number of molecules, which is $0.01\,\text{mol}$ ions.

Hello,

In case of $MgSO_4$ we get 0.02 mol $Mg SO_4$ molecules. So, we get $Mg^{2+}$ and $SO_4^{2-}$ ions. so the number of ions are 0.04 mol.

In case of RbCl , we get 0.02 mol RbCl molecules. So we get $Rb^{1+}$ and $Cl^{1-}$ ions. So the number of ions are 0.04 mol.

In case of $CaCl_2$. we get 0.02 mol $CaCl_2$ molecules. So, we get $Ca^{2+}$ ion and 2 ions of $Cl^{1-}$. So, the number of ions are 0.06 mol.So, answer to this question is 4)