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Calculate the molarity of each of the following solutions:
(a) \(30 \mathrm{~g}\) of \(\mathrm{Co}\left(\mathrm{NO}_3\right)_2 .6 \mathrm{H}_2 \mathrm{O}\) in \(4.3 \mathrm{~L}\) of solution (b) \(30 \mathrm{~mL}\) of \(0.5 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\) diluted to \(500 \mathrm{~mL}\).
(a) \(30 \mathrm{~g}\) of \(\mathrm{Co}\left(\mathrm{NO}_3\right)_2 .6 \mathrm{H}_2 \mathrm{O}\) in \(4.3 \mathrm{~L}\) of solution (b) \(30 \mathrm{~mL}\) of \(0.5 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\) diluted to \(500 \mathrm{~mL}\).
Solution:
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Verified Answer
(a) Molar mass of
\(\begin{aligned}
\mathrm{Co}\left(\mathrm{NO}_3\right)_2 \cdot 6 \mathrm{H}_2 \mathrm{O} &=59+2(14+3 \times 16)+6 \times 18 \\
&=291
\end{aligned}\)
Number of moles of \(\mathrm{Co}\left(\mathrm{NO}_3\right)_2 \cdot 6 \mathrm{H}_2 \mathrm{O}=\frac{30}{291}=0.103\) Vol. of solution \(=4.3 \mathrm{~L}\)
\(\therefore \quad\) Molarity \(=\frac{0.103}{4.3}=0.023 \mathrm{M}\)
(b) \(1000 \mathrm{~mL}\) of \(0.5 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\) contain \(\mathrm{H}_2 \mathrm{SO}_4\) \(=0.5 \mathrm{~mole}\)
\(\therefore 30 \mathrm{~mL}\) of \(0.5 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\) contain \(\mathrm{H}_2 \mathrm{SO}_4\)
\(=\frac{0.5}{1000} \times 30=0.015 \mathrm{~mole}\)
Volume of solution \(=500 \mathrm{~mL}=0.5 \mathrm{~L}\)
\(\therefore \quad\) Molarity \(=\frac{0.015}{0.5}=0.03 \mathrm{M}\)
\(\begin{aligned}
\mathrm{Co}\left(\mathrm{NO}_3\right)_2 \cdot 6 \mathrm{H}_2 \mathrm{O} &=59+2(14+3 \times 16)+6 \times 18 \\
&=291
\end{aligned}\)
Number of moles of \(\mathrm{Co}\left(\mathrm{NO}_3\right)_2 \cdot 6 \mathrm{H}_2 \mathrm{O}=\frac{30}{291}=0.103\) Vol. of solution \(=4.3 \mathrm{~L}\)
\(\therefore \quad\) Molarity \(=\frac{0.103}{4.3}=0.023 \mathrm{M}\)
(b) \(1000 \mathrm{~mL}\) of \(0.5 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\) contain \(\mathrm{H}_2 \mathrm{SO}_4\) \(=0.5 \mathrm{~mole}\)
\(\therefore 30 \mathrm{~mL}\) of \(0.5 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\) contain \(\mathrm{H}_2 \mathrm{SO}_4\)
\(=\frac{0.5}{1000} \times 30=0.015 \mathrm{~mole}\)
Volume of solution \(=500 \mathrm{~mL}=0.5 \mathrm{~L}\)
\(\therefore \quad\) Molarity \(=\frac{0.015}{0.5}=0.03 \mathrm{M}\)
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