Question

Calculate the maximum volume of $0.25M$ copper (II) nitrate solution that can be prepared by dissolving $36.25$ g of copper (II) nitrate trihydrate in sufficient water. Give answer in decilitre.

• A. $8$
• B. $7$
• C. $6$
• D. $9$

Answer 1

The correct option is C $6$

Molar mass of copper (II) nitrate trihydrate $=241.5$ g/mol.
Moles of copper (II) nitrate trihydrate $=\frac{36.25}{241.5}=\frac{3}{20}$.
Therefore, $0.25=\frac{\frac{3}{20}}{V}$

or, $V=\frac{3\times 4}{20\times 1}=\frac{3}{5}=0.6L$ $=6$ decilitre.

Hence, option $C$ is correct.