Answers

Explaination

The questions in this section build upon the last section. They are slightly more complicated, but if you remember the units in each case they will help to tell you what to do. And keep track of the factors of 10. Here are a couple of worked examples.

Q 47.     The relative molecular mass of lactic acid is 90. Therefore the amount of lactic acid in 10 ml of a solution containing 0.9 g of lactic acid per litre is:

Using the method worked through in the previous section
If 0.9g = N moles
90g x N = 0.9g and from this N = 10^-2 moles
thus a solution of 0.9 g l^-1 has a concentration of 10^-2 mol.l^-1 (10 mmol l^-1 )

And remember that 10 ml = 10 × 10^-3 l

So you have worked out a concentration and a volume, and as
concentration = amount ÷ volume
amount = concentration ÷ volume

amount = (10 × 10^-3) mol l^-1 = (10^-2) mol
                (10 × 10^-3) l             (10^-2)

amount = 10^-2 × 10^-2 mol = 10^-4 mol = 0.1 mmol (answer 2)

Q 48.     follows the same method. The relative molecular mass of ethanal is 45. Therefore the amount of ethanal in 100 ml of a solution containing 0.9 g of ethanal per litre is:
Concentration of solution = (0.9 × 45) moles per litre = 0.02 mol l^-1

Now, if you remember the equation concentraton = amount / volume  then amount = concentration x volume
The concentration is always given in moles per litre (mol l^-1), so you have to get the volume into litres too.
Volume = 100 ml = 100 × 10^-3 l = 10^-1 l = 0.1 l
Amount = 0.02 mol l^-1 x 0.1 l = 0.002 mol = 2 mmol (answer 3)
Note that the litre part of the units cancel each other out.
 

Q 50.     The amount in 1 ml of the solution is actually 10 m mol, which is "none of the above"; answer 5.