Boiling-point elevation is a colligative property.
That means, the the boiling-point elevation depends on the molar content (fraction) of solute.
The dependency is ΔTb = Kb*m
Where ΔTb is the elevation in the boiling point, kb is the boiling constant, and m is the molality.
A solution of 6.00 g of Ca(NO3) in 30.0 g of water has 4 times the molal concentration of a solution of 3.00 g of Ca(NO3)2 in 60.0 g of water.:
(6.00g/molar mass) / 0.030kg = 200 /molar mass
(3.00g/molar mass) / 0.060kg = 50/molar mass
=> 200 / 50 = 4.
Then, given the direct proportion of the elevation of the boiling point with the molal concentration, the solution of 6.00 g of CaNO3 in 30 g of water will exhibit a greater boiling point elevation.
Or, what is the same, the solution with higher molality will have the higher boiling point.
Answer:
can you help mine please
How many molecules of chlorine are needed to react with 56.Og of iron to form Iron (III) chloride (FeCl3)?
You first add the manganese and exchange the number of electrons needed with the hydroxide. While the hydroxide needs only 1 electron the manganese needs 4, so after you exchange the electrons the manganese will be just 1 atom while the hydroxide is 4. Mn(OH)4
