Answer:
[tex]\rm pOH \approx 1.279[/tex] if [tex]2.95\; \rm g[/tex] of [tex]\rm KOH[/tex] is dissolved completely in water to make a [tex]100\; \rm mL[/tex] solution.
Explanation:
Look up the relative atomic mass data on a modern periodic table:
Calculate the formula mass of [tex]\rm KOH[/tex]:
[tex]\begin{aligned}&M(\mathrm{KOH})\\ &= 39.098 + 15.999 + 1.008 \\ &= 56.105\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
Find the number of moles of formula units in that [tex]2.95\; \rm g[/tex] of [tex]\rm KOH[/tex]:
[tex]\begin{aligned}& n(\mathrm{KOH}) \\ & = \frac{m(\mathrm{KOH})}{M(\mathrm{KOH})} = \frac{2.95\; \rm g}{56.105\; \rm g \cdot mol^{-1}} \approx 0.0525800\; \rm mol \end{aligned}[/tex].
[tex]\rm KOH[/tex] is a strong base. When dissolved in water, it ionizes completely to produce [tex]\rm K^{+}[/tex] ions and hydroxide ions ([tex]\rm OH^{-}[/tex].)
Note that there are one moles of [tex]\rm OH^{-}[/tex] ions in each mole of [tex]\rm KOH[/tex] formula units. When that [tex]2.95\; \rm g[/tex] of [tex]\rm KOH[/tex] (approximately [tex]0.0525800\; \rm mol[/tex] of [tex]\rm KOH[/tex] formula units) dissolves completely in water to make a [tex]100\; \rm mL[/tex] solution, about [tex]0.0525800\; \rm mol[/tex] of [tex]\rm OH^{-}[/tex] ions will be produced.
Convert the unit of volume to liters: [tex]V = 100\; \rm mL = 0.1\; \rm L[/tex].
Calculate the concentration of [tex]\rm OH^{-}[/tex] ions in that solution:
[tex]\begin{aligned}&[\mathrm{OH^{-}}] \\ &= \frac{n}{V} \approx \frac{0.0525800\; \rm mol}{0.1 \; \rm L} \approx 0.525800\; \rm mol \cdot L^{-1} \end{aligned}[/tex].
Calculate the [tex]\rm pOH[/tex] of that solution:
[tex]\begin{aligned}& \mathrm{pOH} \\ &= -\log_{10}[\mathrm{OH^{-}}] \\ &\approx -\log_{10}(0.525800) \approx 1.279\end{aligned}[/tex].
