Asolution is made by dissolving 1.00 m of sodium chloride (nacl) in 155 grams of water. if the molar boiling point constant for water (kb) is 0.51 c/m what would be the boiling point of this solution?

106.58°C

1) Data

- solute: NaCl
- n solute = 1.00 moles
- solvent: water
- grams of solvent = 155 g
- kb = 0.51 °C / m
- Tb ?

2) Formulas

- ΔTb = Tb - normal Tb
- ΔTb = i * kb * m
- m = moles of solute / kg of solvent

3) Solution:

3.1) m = n solute / kg solvent = 1.00 moles / 0.155 kg = 6.4516 m

3.2) i for NaCl is 2, because each molecule of NaCl dissolves into 2 ions.

3.3) ΔTb = 2 * 0.51 °C/m * 6.4516m = 6.58°C

3.3) ΔTb = Tb - normal Tb => Tb = ΔTb + normal Tb

normal Tb of water = 100°C

Tb = 6.58 °C + 100 °C = 106.58°C

106.58°C

Most likely a is the answer - because  you generate   38 atp's but also consume an atp.   each atp gives approx. less than 4 kcal so probably a is the closest answer.