calculate the molarity of na2co3 solution prepared by dissolving 5.2g in enough water to make 500 ml of the solution molecular mass of na2co3 in 106g
Answers
What the question is asking is the molar concentration of the given substance after dilution with the help of values given. The first step of the answer is converting the given weight of Na2CO3⋅10H2O into amount of substance. For that we have a formula as
amount of substanceamount of substance=mass of substancemolecular mass of the substance=1 g Na2CO3⋅10H2O286 g Na2CO3⋅10H2O=0.003447 mol.
We used this first step to determine the amount of substance of Na2CO3⋅10H2O present in the solution. It is necessary because even after dilution of the solution the amount of substance will remain constant.
The second step is converting the amount of substance into molarity. As,
molaritymolarity=amount of substancevolume in L=0.003447 mol0.020 L=0.174825 M
The second step is done to find out the concentration of the substance in terms of molarity since the question has asked us to give the answer in molarity after dilution.
After that the given solution is diluted up to 250 mL, so we have to use dilution formula
initial concentration×initial volume of solution=final concentration×final volume of solution.
So we suppose that the final concentration of the solution is x and we have values for others entities.
(0.174825 M)(0.020 L)x=(0.250 L)(x)=0.013986 M
This last step is carried out because after dilution the molarity of the solution changes but not the amount of substance as I stated earlier. So we equate the equation as
initial amount of substance=final amount of substance.
And the final answer is 0.013986 M.
Answer:
Here is what I did and got the answer and don't understand why I need to do it:
1 g Na2CO3⋅10H2O286 g Na2CO3⋅10H2O0.003447 mol0.020 L(0.174825 M)(0.020 L)x=0.003447 mol=0.174825 M=(0.250 L)(x)=0.013986 M