Question

A $15$ mL sample of $0.20MMgC{l}_2$ is added to $45$ mL of $0.40MAlC{l}_3$, what is the molarity of $C{l}^{-}$ions in the final solution?

• A. $1.0M$
• B. $0.60M$
• C. $0.35M$
• D. $0.30M$

Answer 1

The correct option is A $1.0M$

The number of mmol of $MgC{l}_2$ present $=15\times 0.20=3$ mmol.

The number of mmol of $AlC{l}_3$ present $=45\times 0.40=18$ mmol.

The number of mmol of $C{l}^{-}$ present $=(2\times 3)+(3\times 18)=60$ mmol.

Total volume $=15+45=60$ mL

The molarity of chloride ions in the final solution $=\frac{60}{60}=1.0M$.

Hence, the correct option is $A$