I have to make a 250 mL solution of $\ce{Mg(NO3)2}$, the concentration of the solution has to be 0.3 mol/L. How can I find out the mass of how much $\ce{Mg(NO3)2}$ I need to use?


In 1 L there are 1000 mL and according to question there is 0.3 mol in 1000 mL. So, we can calculate how many moles should be there in 250 mL, i.e. (250/1000)*0.3 = 0.075 moles.

Now, let's find out the molecular mass of your compound i.e. 24 + 14*2 +16*6 = 148 u. So, by mole concept we know that there will be 148 g in 1 mole of your compound but you have 0.075 moles.

So the mass needed is (0.075/1)*148 = 11.1

Hope that makes sense !


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