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

Step-by-step explanation:

Answer 2

Approximately
`0.821\; \rm mol`.

Step-by-step 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*(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})} \\ &= (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})} \\ &= (1.37159\; \rm mol)/(1.67\; \rm mol \cdot L^(-1)) \approx 0.821\; \rm L \end{aligned}`.