00:01
Hi, so to solve for this we will use the formula for molarity.
00:05
Molarity is equivalent to the number of moles of the solute over the volume of the solution in terms of liters.
00:14
So the first question is the molarity of the solution.
00:19
So aluminum nitrate solution that's alno3.
00:25
If we have 24 .9 grams of alno3, convert this to moles by dividing the mass of this compound that will be 213 grams and then we have here moles over the volume of the solution we have 500 ml.
00:46
We have to convert ml to liters.
00:48
1000 ml is 1 liter.
00:51
So that means you could cancel grams of aluminum nitrate ml ml and then we have here the remaining units.
00:56
We have moles per liter.
00:59
Solving for this one, this will give us 0 .234.
01:04
This is moles of aluminum nitrate per liter solution or molar.
01:13
That's the answer to the first question.
01:15
The second question is concentration of the aluminum cation...