Question

$50mL$ of $0.20M$ aluminum bromide, $AlB{r}_3$, is added to $150mL$ of $0.20MMgB{r}_2$.
Which of the following choices CORRETLY gives the concentration of bromide ion in the resulting solution?

• A. $0.45M$
• B. $0.20M$
• C. $0.90M$
• D. $0.55M$

Answer 1

The correct option is A $0.45M$

No. of millimoles of bromine ions in $50mL$ of $0.2MAlB{r}_3=(50\times 0.2)\times 3\times {10}^{-3}moles=3\times {10}^{-2}moles$

No. of millimoles of bromine ions in $150mL$ of $0.2MMgB{r}_2=2\times (150\times 0.2)\times {10}^{-3}moles=6\times {10}^{-2}moles$

So, the total no. of moles $=9\times {10}^{-2}moles$

So, volume of bromine ions is $200\space mL.$

$\Rightarrow$ Molarity of Bromine ions $=\text{No. of moles}\times \frac{1000}{V\left(mL\right)}=(9\times {10}^{-2})\times 5=45\times {10}^{-2}=0.45M$