Answer : The original concentration of copper (II) sulfate in the sample is, 
Explanation :
Molar mass of Cu = 63.5 g/mol
First we have to calculate the number of moles of Cu.
Number of moles of Cu = 
Now we have to calculate the number of moles of 
Number of moles of Cu = Number of moles of 
Number of moles of
= 
Now we have to calculate the molarity of 

Now put all the given values in this formula, we get:

To change mol/L into g/L, we need to multiply it with molar mass of 
Molar mass of
= 159.609 g/mL
Concentration in g/L = 
Thus, the original concentration of copper (II) sulfate in the sample is, 
Answer:
5
Explanation:
Firstly, we convert what we have to percentage compositions.
There are two parts in the molecule, the sulphate part and the water part.
The percentage compositions is as follows:
Sulphate- (103.74)/(103.74 + 58.55) × 100% = apprx 64%
The water part = 100 - 64 = 36%
Now, we divide the percentages by the molar masses.
For the CuSO4 molar mass is 64 + 32 + 4(16) = 160g/mol
For the H2O = 2(1) + 16 = 18g/mol
Now we divide the percentages by these masses
Sulphate = 64/160 = 0.4
Water = 36/18 = 2
The ratio is thus 0.4:2 = 1:5
Hence, there are 5 water molecules.