Answer:
Approximately [tex]0.821\; \rm mol[/tex].
Explanation:
Look up the relative atomic mass of [tex]\rm Ba[/tex], [tex]\rm O[/tex], and [tex]\rm H[/tex] on a modern periodic table:
Calculate the formula mass of [tex]{\rm Ba(OH)_2}[/tex]:
[tex]\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}[/tex].
Calculate the number of moles of [tex]{\rm Ba(OH)_2}[/tex] formula units in that [tex]235\; \rm g[/tex] of this compound:
[tex]\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}[/tex].
Calculate the volume of a [tex]c({\rm Ba(OH)_2}) = 1.67\; \rm mol \cdot L^{-1}[/tex] with approximately [tex]n({\rm Ba(OH)_2}) = 1.37159\; \rm mol[/tex] of the solute:
[tex]\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}[/tex].
