How to calculate the new concentration of a solution after adding more solute and convert it to ppm? My question is the following: 


  $2~\mathrm{g}$ table salt is added to $0.5~\mathrm{m^3}$ water whose salt concentration is $10~\mathrm{mg\,L^{-1}}$. Compute the salt concentration of the mixture in ppm. 


I am fairly new to chemistry actually this is my first time taking a chemistry course of this level so I'm having quite a hard time. Therefore, if anyone could recommend some websites where I could receive help or even videos, such as khanacademy that would be greatly appreciated! 

As for the following question all I know is that $2~\mathrm{g} = 2000~\mathrm{mg}$
and ppm is calculated by: 

$$\text{ppm} = \frac{\text{grams of solute}}{\text{grams of solution}}\times  1000000 $$

However, im confused as to how much grams of solute and grams of solution there will be in the mixture after adding the $2~\mathrm{g}$ of salt?
Let's think about it :

You have the formula of the mass concentration $C_m$: 

$C_m = \frac{m}{V}$, 

where $m$ is the mass in grams of solute and $V$ the volume in liters of the solution. 

A first step for you would be to calculate the mass of solute already in the solution : your solution has a volume of $0.5~\mathrm m^3$, which is $V=500~\mathrm L$, with a massic concentration of $C_m=10~\mathrm{mg\cdot L^{-1}}$. Therefore, there is already a mass 

$m=C_m~\times~V = 10\cdot10^{-3} \times 500 = 5~\mathrm g$ 

of salt in your solution. 

To this mass, you add $2~\mathrm g$ of salt, so the mass of solute in your solution is now $m=5~\mathrm g + 2~\mathrm g = 7~\mathrm g$.

So, at end you have a mass $m=7~\mathrm g$ of salt. 

I let you compute the mass of solution, knowing that $1~\mathrm L$ of water weights $1~\mathrm{kg}$.