Calculate the boiling point of a 3.71 mol aqueous calcium bromide solution. All right, so they want us to find our boiling point. So remember, your boiling point of solution equals the boiling point of your pure solvent plus the change in the boiling point.
So the boiling point of our pure salt of our pure solvent. What exactly is our solvent here? Well, here they're saying aqueous. When we say aqueous that means the solvent is water. Pure water boils at 100℃. All we have to do now is figure out what our new changing boiling point will be.
So we're going to say ΔTB equals I times KB times M. Remember, I is the number of ions your solute will dissolve into when placed in a solvent. Here, calcium bromide is ionic. It breaks up into a calcium ion +2 bromide ions, for a total of 3 ions. The solvent is water. The boiling point constant of water is oh, .51℃ over molality.
Then finally we're going to say our molality has given us given to us as 3.71 mole. So molalodies cancel out and I'll have my answer here as degrees Celsius, which comes out to 5.68℃. So we take that and we plug it here. So when we add those together, our new temperature is 105.68℃ Celsius.
Remember, it's called boiling point elevation. We go up as we add solute to our pure solvent. Here pure water was at 100℃. After the addition of calcium bromide, it's now at 105.68℃.