Actually, i want to know, how to calculate the amount (gram) of Pb required from Pb(NO3)2 to make a 1000 ppm Pb solution, not 1000 ppm Pb(NO3)2 solution. 

Thanks a lot for your co-operation.

I just want to simplify the calculation and I hope it will help understanding it better. To prepare 1 mg of Pb you need 1.6 mg of Pb(NO3)2 also to make 1 g of Pb you need 1.6 g of (Pb(NO3)2)

To obtain 1000mg of lead from lead nitrate you need 1598 mg of lead nitrate dissolve in 1L. That is approximately 1.6 g of lead nitrate.

(331.3g/ mol x 1000mg lead)/207.2 g/mol= 1598.9 mg lead nitrate

Molar mass lead=207.2 g/mol

Molar mass lead nitrate=331.2g/mol

Why is it that, after testing the concentration of the solution prepared does not give exactly 50 ppm. Instead it gives a higher concentration??

All you need is to know the molecular weight of Pb(NO3)2 for example its X and you need to now the molecular weight of lead for example its Y then u divide Y/X . the amount u will get should be dissolved in 1000 ml of distilled water will be your 1000ppm conc.