Find out the molarity of $30\mathrm{g}$ of $\mathrm{Co}{\left({\mathrm{NO}}_3\right)}_2.6{\mathrm{H}}_2\mathrm{O}$ in $4.3\mathrm{L}$ of solution.

Part-$1$ Given data:

1. Mass of compound $=30\mathrm{g}$
2. Volume of solution $=4.3\mathrm{L}$

Part-$2$: Calculate the molar mass$\left(\mathrm{M}\right)$ of $\mathrm{Co}{\left({\mathrm{NO}}_3\right)}_2.6{\mathrm{H}}_2\mathrm{O}$

$\mathrm{M}=58.93+(14\times 2)+(16\times 6)+(6\times 2)+(6\times 16)=290.93{\mathrm{gmol}}^{-1}$.

Part-$3$:Calculate number of moles$\left(\mathrm{n}\right)$

Number of moles $=$$\frac{Givenmass\left(m\right)}{Molarmass\left(M\right)}$$=\frac{30}{290.93}=0.103mol$

Part-$4$: Calculate molarity$\left(\mathrm{M}\right)$

1. Molarity is defined as the number of moles of solute per litre of solution.
2. $M=\frac{0.103}{4.3}=0.024M$