A 4.00 g sample of potassium hydroxide (KOH, $M=56.11$ g mol${}^{-1}$) is dissolved in 300.0 g of water. Assuming KOH dissociates completely into two ions, calculate the boiling point elevation ($Delta{T}_b$) of the resulting solution. Use the ebullioscopic constant of water $K_b=0.512$ K kg mol${}^{-1}$.

Step-by-step solution

1. Calculate the number of moles of KOH: $n=\frac{4.00\text{g}}{56.11{\text{g,mol}}^{-1}}=0.0713\text{mol}$.
2. Determine the number of moles of solute particles (ions): $i=2$, so $n_{\text{particles}}=i,n=2(0.0713)=0.1426\text{mol}$.
3. Compute the molality of solute particles: $m_{\text{particles}}=\frac{n_{\text{particles}}}{\text{kg of solvent}}=\frac{0.1426}{0.300}=0.475{\text{mol,kg}}^{-1}$.
4. Use the formula for boiling‑point elevation for a dilute solution: $\Delta T_b=K_b,m_{\text{particles}}=0.512,(0.475)=0.2435\text{K}$.
5. Round to two significant figures: $\Delta T_b=0.24\text{K}$.

Concept tested: Boiling‑point elevation

Published 10 October 2026 · How this solution was prepared and checked