Question

1. instant ramen seasoning packet

the seasoning packet effectively dissolves 8.0 g of salt (nacl) into 500 g of boiling water. (assume i=2, kb=0.512)
what temperature will the solution boil at?

Explanation:

Step1: Convert mass of water to kg

Mass of water = 500 g = 0.5 kg

Step2: Calculate moles of NaCl

Molar mass of NaCl ≈ 58.5 g/mol
Moles of NaCl = $\frac{8.0\ \text{g}}{58.5\ \text{g/mol}} ≈ 0.1368\ \text{mol}$

Step3: Find molality of NaCl

Molality ($m$) = $\frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{0.1368\ \text{mol}}{0.5\ \text{kg}} ≈ 0.2736\ \text{mol/kg}$

Step4: Calculate boiling point elevation

$\Delta T_b = i \times K_b \times m = 2 \times 0.512\ ^\circ\text{C·kg/mol} \times 0.2736\ \text{mol/kg} ≈ 0.281\ ^\circ\text{C}$

Step5: Compute boiling point of solution

Boiling point = $100^\circ\text{C} + 0.281^\circ\text{C} ≈ 100.28^\circ\text{C}$

Answer:

100.28°C