Question

1. saltwater for cooking pasta ≈0.598℃ a chef adds 25.0 g of nacl to 1.00 kg of water. - assume i=2 what is the boiling point of the salty water? 4. radiator fluid (boiling protection) a car radiator contains 3.00 kg of water with 400 g ethylene glycol (c₂h₆o₂) added (non - electrolyte).

Explanation:

Step1: Calculate moles of NaCl

Molar mass of NaCl = 58.44 g/mol. Moles = $\frac{25.0\ \text{g}}{58.44\ \text{g/mol}} \approx 0.4278\ \text{mol}$

Step2: Find molality of solution

Mass of water = 1.00 kg. Molality ($m$) = $\frac{0.4278\ \text{mol}}{1.00\ \text{kg}} = 0.4278\ \text{mol/kg}$

Step3: Calculate boiling point elevation

$\Delta T_b = i \cdot K_b \cdot m$. $i=2$, $K_b(\text{water})=0.512^\circ\text{C·kg/mol}$. $\Delta T_b = 2 \times 0.512 \times 0.4278 \approx 0.437^\circ\text{C}$

Step4: Find new boiling point

Pure water boils at $100^\circ\text{C}$. New boiling point = $100 + 0.437 \approx 100.44^\circ\text{C}$

Answer:

100.44°C