Can someone talk me through this problem and explain the process?

A solution is made by dissolving 1.00 moles of sodium chloride (NaCl) in 155 grams of water. If the molal boiling point constant for water (Kb) is 0.51 °C/m, what would be the boiling point of this solution?

Question

Can someone talk me through this problem and explain the process?

A solution is made by dissolving 1.00 moles of sodium chloride (NaCl) in 155 grams of water. If the molal boiling point constant for water (Kb) is 0.51 °C/m, what would be the boiling point of this solution?

aaronq · Answer

1. Find the molality

         $m=\dfrac{n_{solute}}{kg_{solvent}}$

2. substitute the values given into:

         $\Delta T =i*m*K_b$

i = van't hoff factor; approximated by the number of molecules/atoms present upon dissociation of species in the solvent.

             NaCl (s) -> $Na^+_{\;aq.} +Cl^-_{\;aq.}$    i=2

$\Delta T$ = change in temperature

m=molality

Kb= boiling point elevation constant

After you found $\Delta T$, you add it to the temp at which the solvent boils. Water boils at 100 Celsius, so, for this question:

Boiling Temp of solution = $100 C^{\;o} + \Delta T$

anonymous · Answer

Thank you! This helped me so much!

aaronq · Answer

no problem, glad i could help !