Difficult
Calculate the molarity of each of the following solutions: $(a)$ $30 \, g$ of $Co(NO_3)_2 \cdot 6H_2O$ in $4.3 \, L$ of solution $(b)$ $30 \, mL$ of $0.5 \, M \, H_2SO_4$ diluted to $500 \, mL$.
(N/A) Molarity is defined as the number of moles of solute per liter of solution: $M = \frac{n}{V(L)}$.
$(a)$ Molar mass of $Co(NO_3)_2 \cdot 6H_2O = 59 + 2(14 + 48) + 6(18) = 291 \, g \, mol^{-1}$.
Moles of $Co(NO_3)_2 \cdot 6H_2O = \frac{30 \, g}{291 \, g \, mol^{-1}} \approx 0.103 \, mol$.
Molarity $= \frac{0.103 \, mol}{4.3 \, L} \approx 0.0239 \, M$.
$(b)$ Using the dilution formula $M_1V_1 = M_2V_2$:
$0.5 \, M \times 30 \, mL = M_2 \times 500 \, mL$.
$M_2 = \frac{0.5 \times 30}{500} = 0.03 \, M$.
$(a)$ Molar mass of $Co(NO_3)_2 \cdot 6H_2O = 59 + 2(14 + 48) + 6(18) = 291 \, g \, mol^{-1}$.
Moles of $Co(NO_3)_2 \cdot 6H_2O = \frac{30 \, g}{291 \, g \, mol^{-1}} \approx 0.103 \, mol$.
Molarity $= \frac{0.103 \, mol}{4.3 \, L} \approx 0.0239 \, M$.
$(b)$ Using the dilution formula $M_1V_1 = M_2V_2$:
$0.5 \, M \times 30 \, mL = M_2 \times 500 \, mL$.
$M_2 = \frac{0.5 \times 30}{500} = 0.03 \, M$.
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