Question

How many liters of water are needed to prepare a 1.67M solution of Ba(OH)2 if you need to dissolve 235g of it?

Answer 1

Answer: 2.04

Explanation:

Answer 2

Answer:

Approximately $0.821\; \rm mol$.

Explanation:

Look up the relative atomic mass of $\rm Ba$, $\rm O$, and $\rm H$ on a modern periodic table:

• $\rm Ba$: $137.327$.
• $\rm O$: $15.999$.
• $\rm H$: $1.008$.

Calculate the formula mass of ${\rm Ba(OH)_2}$:

$\begin{aligned}& M({\rm Ba(OH)_2}) \\ &= 137.327 + 2\times(15.999 + 1.008) \\ &\approx 171.334\; \rm g \cdot mol^{-1}\end{aligned}$.

Calculate the number of moles of ${\rm Ba(OH)_2}$ formula units in that $235\; \rm g$ of this compound:

$\begin{aligned}& n({\rm Ba(OH)_2}) \\ &= \frac{m({\rm Ba(OH)_2})}{M({\rm Ba(OH)_2})} \\ &= \frac{235\; \rm g}{171.334\; \rm g \cdot mol^{-1}} \approx 1.37159\; \rm mol \end{aligned}$.

Calculate the volume of a $c({\rm Ba(OH)_2}) = 1.67\; \rm mol \cdot L^{-1}$ with approximately $n({\rm Ba(OH)_2}) = 1.37159\; \rm mol$ of the solute:

$\begin{aligned}& V({\rm Ba(OH)_2}) \\ &= \frac{n({\rm Ba(OH)_2})}{c({\rm Ba(OH)_2})} \\ &= \frac{1.37159\; \rm mol}{1.67\; \rm mol \cdot L^{-1}} \approx 0.821\; \rm L \end{aligned}$.