(a) 30 g of Co(NO)3. 6H2O in 4.3 L of solution
(b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.
We are aware of this.,
Molarity $=\frac{\text { Moles of Solute }}{\text { Volume of solution in litre }}$
(a) Molar mass of $\mathrm{Co}(\mathrm{NO})_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}=59+2(14+3 \times 16)+6 \times 18=291 \mathrm{~g} \mathrm{~mol}^{-1}$
so, Moles of $\mathrm{Co}(\mathrm{NO})_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}=\frac{30}{291} \mathrm{~mol}$
$=0.103 \mathrm{~mol}$
so, molarity $=\frac{0.103 \mathrm{~mol}}{4.3 L}$
$=0.023 \mathrm{M}$
(b) Total number of moles in $1000 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}=0.5 \mathrm{~mol}$
So, Number of moles found in $30 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}=\frac{0.5 \times 30}{1000} \mathrm{~mol}$
$=0.015 \mathrm{~mol}$
Hence, molarity $=\frac{0.015}{0.5 L}$ mol $=0.03 \mathrm{M}$