What is the difference between weight and mass?
2 answers:
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There are several various of expressing concentration. For instance, mass percent, volume percent, Molarity, Normality, Molality, etc.
In present case, weight of solute and solvent are given, so it will be convenient to express concentration in terms of mass percent.
Given: weight of solute (Ca2+) = 8500 g
weight of solvent (water) = 490 g.
Therefore, mass of solution = 8500 + 490 = 8990 g
Now, mass percent =
=
= 94.55 %
Answer: Concentration of calcium ions is in this solution is 94.55 % (w/W)
In present case, weight of solute and solvent are given, so it will be convenient to express concentration in terms of mass percent.
Given: weight of solute (Ca2+) = 8500 g
weight of solvent (water) = 490 g.
Therefore, mass of solution = 8500 + 490 = 8990 g
Now, mass percent =
=
= 94.55 %
Answer: Concentration of calcium ions is in this solution is 94.55 % (w/W)
1.66 M is the concentration of the chemist's working solution.
<h3>What is molarity?</h3>Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dilute solution of this reactant.
The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.
We want to know the concentration of this diluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:
=
(1)
and we also know that:
n = M x
If we replace this expression in (1) we have:
x
=
x
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is :
4.93 x 210 = 620 x
= 1.66 M
This is the concentration of the solution prepared.
Learn more about molarity here:
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