00:03
Here in this problem, given a solution contains 35 .0 gram ammonia in 75 .0 gram of water.
00:12
Density of the solution given here, we have to calculate molarity and molality of the solution.
00:20
Now, mass of ammonia is given here.
00:24
So we can find out moles of ammonia.
00:27
So moles of nh3, that is the mass given.
00:33
Which is 35 .0 gram divided by the molar mass of ammonia that is 17 .031 gram per mole.
00:47
And we get 2 .055 moles of nhc.
00:55
Now, here mass of solution, 35 gram plus 75 gram sulfate.
01:03
So, mass of solution here, mass of solution, mass of salute, that is 35 .0 gram ammonia and 75 gram water here.
01:19
So, let's add them together, 75 .0 grams.
01:26
So we get 110 gram of mass of solution.
01:31
Now density of solution given, so we can find out volume of the solution.
01:39
Volume of the solution, the formula is volume is volume is mass divided by density.
01:46
Mass is 100 ,000 gram divided by the density of the solution, which is this one.
01:53
1 .83 gram per milliliter.
01:59
Gram per milliliter.
02:01
And we get volume of the solution.
02:03
Solution 60 .1093 milliliter.
02:09
Now we'll convert it into liter.
02:12
So 60 .11 times 10 to the power negative 3 liter...