Calculate the molarity of each of the following solutions:

$A$: $30g\text{of}Co(N{O}_3{)}_2.6H_{2}O\text{in}4.3L$ of solution.

$B$: $30mL\text{of}0.5M{H}_{2}S{O}_4$ diluted to $500mL$.

$A$:

Given:
Mass of solute $\left(w_2\right)=30g$
Volume of solution $=4.3L$
Molar mass of given solute $CO(N{O}_3{)}_2.6H_{2}O$
$=58.7+2[14+(16\times 3\left)\right]+6(2+16)$
$=58.7+(2\times 62)+(6\times 18)$
$=58.1+124+108$
$=290.7gmo{l}^{-1}$

Moles of $CO(N{O}_3{)}_2.6H_{2}O$

Number of moles of $CO(N{O}_3{)}_2.6H_{2}O=\frac{\left(\text{Msss of cobalt nitrate}\right)}{\left(\text{Molar mass of cobalt nitrate}\right)}$

Number of moles of $CO(N{O}_3{)}_2.6H_{2}O=\frac{30}{290.7}=0.103mol$

Molarity of $CO(N{O}_3{)}_2.6H_{2}O$

Molarity $\left(M\right)=\frac{\text{moles of solute}}{\text{Volume of solution (L)}}$

Molarity $\left(M\right)=\frac{0.103}{4.3}=0.0239M$

Final Answer:

The molarity of
$30g\text{of}Co(N{O}_3{)}_2.6H_{2}O$ in $4.3L$ of solution is $0.0239M$.

$B$:

Given:

$30mL$ of $0.5M{H}_{2}S{O}_4$ dilutes to $500mL$

Molarity of Final mixture

For dilution from volume $V_{1}to{V}_2$

$M_1{V}_1=M_2{V}_2$

putting the values,
$V_1=30mL$
$M_1=0.5M$
$V_2=500mL$

We get,

$M_2=\frac{30\times 0.5}{500}=0.03M$
Thus, final molarity $=0.03M$

The molarity of $30mL\text{of}0.5M{H}_{2}S{O}_4$ diluted to $500mL\text{is}0.03M$.