Answer:
What is the Boiling Point of the solution resulted from the dissolving of 32.5 g of NaCl in 250.0 g of water? 102,31°C
Explanation:
This question involves the Elevation of boiling point
ΔT = Kb . m . i
(T°solution - T°solvent pure) = Ebulloscopic constant . molality . Van 't Hoff factor
Van 't Hoff factor is 2 for NaCl because you have two ions on disociation.
Kb for water is, 0,52 °C . kg/mol - It is a known value.
You know that water pure boils at 100°C so let's build the formula
(T°solution - 100°C) = 0,52 °C.kg/mol . molality . 2
We have to find out the molality (moles of solute/1kg solvent)
In 250 g H2O we have 32,5 g NaCl so the rule of three will be
250 g H2O ______32,5 g NaCl
1000g __________ (1000g . 32,5g) / 250g = 130 g NaCl
Molar mass NaCl : 58,45 g/m
Moles NaCl : mass/ molar mass --> 130g /58,45 g/m = 2,22 m
(T°solution - 100°C) = 0,52 °C.kg/mol . 2,22 mol/gk . 2
T°solution - 100°C = 2,31 °C
T° at boiling point in the solution = 2,31°C + 100°C = 102,31°C
