Question

How do I make 100 ml of 0.3 M HCl from 1 M?

Answer 1

Composition of a solution can be expressed based on its concentration. The terms dilute and concentrated provide a vague idea about the concentration of the solution and thus a quantitative representation is needed for expression of the concentration. The concentration of the solution depends on the amount of solute and solvent that are present in the solution.

Complete step by step answer:
Molarity of the solution can be represented in terms of the volume of the solution rather than the mass of the solvent as in case of molality. It can be represented as
${{Molarity (M) = }}\dfrac{{{{Moles\; of \;solute}}}}{{{{Volume\; of \; solution \; in \;litre}}}}$
The moles of the compound can be calculated from the equation given as
$Moles = \dfrac{{mass}}{{molar\; mass}}$
Here we have to use the molarity of the sample given as the target to get to the answer. We need to get the amount of solute contained in the final solution.
Once we get that we can use the molarity of the stock solution to determine the volume of the solution that will contain the moles required of the hydrochloric acid that we need.
So from the above relation, we have the moles of hydrochloric acid as
${n_{HCl}} = 0.3 \times 100 \times {10^{ - 3}} = 0.03$
This is the amount of solute we need to put in the solution
For calculating the volume of the solution we use the above equation as
$Volume = \dfrac{{Moles}}{{Molarity}}$
So we get the Volume as $\dfrac{{0.030}}{1}$ so the volume is 0.03 L that is 30 mL
So for the preparation of the target solution take 30 mL hydrochloric acid and add water to make the volume of solution 100 mL

Note: There are other representations of the concentration of solution than molarity.
While molality is represented by $m$ another representation, Molarity is represented by $M$.
Molarity of the solution can be represented in terms of the volume of the solution rather than the mass of the solvent as I case of molality
For the quantitative representation of the concentration of the solution there are several ways that can be used, molality is one of those criteria.
Molality is represented by $m$ and can be calculated by the following formula
$Molality(m) = \dfrac{{{{Moles \;of \;solute}}}}{{{{Mass\; of\; solvent\; in\; kg}}}}$