A solution is prepared by dissolving 0.87mol of MgCl 2 in 0.40kg of water.
How many moles of ions are present in solution?
What is the change in the boiling point of the aqueous solution?
To find the moles of ions present in the solution, we need to consider the dissociation of magnesium chloride (MgCl2) in water.
The dissociation equation for MgCl2 is:
MgCl2 -> Mg2+ + 2Cl-
From the equation, we can see that one mole of MgCl2 produces two moles of ions.
Given that the solution contains 0.87 moles of MgCl2, we can calculate the number of moles of ions as follows:
Number of moles of ions = 0.87 moles MgCl2 × 2 moles of ions per mole of MgCl2
Number of moles of ions = 1.74 moles of ions
Therefore, there are 1.74 moles of ions present in the solution.
Now, let's move on to calculating the change in the boiling point of the aqueous solution. The change in boiling point is determined by the concentration of the solute and the properties of the solvent.
The change in boiling point (ΔTb) can be calculated using the formula:
ΔTb = Kb × m
Where ΔTb is the change in boiling point, Kb is the molal boiling point elevation constant of the solvent, and m is the molality of the solution.
For water, the molal boiling point elevation constant (Kb) is approximately 0.512°C/m.
To calculate the molality (m) of the solution, we need to find the mass of the solvent (water). Given that the mass of water is 0.40 kg, we can convert it to grams:
Mass of water = 0.40 kg × 1000 g/kg
Mass of water = 400 g
Now, we can calculate the molality (m) of the solution:
m = moles of solute / mass of solvent (in kg)
m = 0.87 mol / 0.400 kg
m = 2.17 mol/kg
Finally, we can calculate the change in boiling point (ΔTb):
ΔTb = 0.512°C/m × 2.17 mol/kg
ΔTb ≈ 1.11°C
Therefore, the change in the boiling point of the aqueous solution is approximately 1.11°C.