What is the molal concentration m of NaCl,a strong electrolyte in water, if the observed boiling temperature of the solution is 100.361⁰C? (Kb = 0.515⁰C/m for water.)
delta T = i*Kb*m
Solve for m; i = 2 for NaCl.
you have to know the initial grams before starting. its not given I have the same question and I can't do it
To determine the molal concentration (m) of NaCl, you can use the formula:
ΔTb = Kb * m
Where:
ΔTb is the change in boiling temperature
Kb is the molal boiling point elevation constant for water
m is the molal concentration of the solute (NaCl)
Given that the observed boiling temperature change (ΔTb) is 100.361⁰C and Kb for water is 0.515⁰C/m, we can rearrange the formula to solve for m:
m = ΔTb / Kb
Substituting the values:
m = 100.361⁰C / 0.515⁰C/m
m ≈ 194.954 mol/kg
Therefore, the molal concentration (m) of NaCl in water is approximately 194.954 mol/kg.