Question

If the density of methanol is $$0.793kg/L$$, what is its volume needed for making $$2.5L$$ of its $$0.25M$$ solution?

Answer 1

Hint: Molarity ($$M$$) indicates the number of moles of solute per litre of solution (moles/Litre) and is one of the most common units used to measure the concentration of a solution. The density of a substance is the relationship between the mass of the substance and how much space it takes up (volume).

Complete answer:
We know that, molarity $$={\displaystyle \frac{\text{moles of solute}}{\text{volume of solution}}}$$
It is given that,
$$0.25M={\displaystyle \frac{n}{2.5L}}$$
Where, $$n\to$$ number of moles
Hence, $$n=0.25M\times 2.5L$$
$$n=0.625moles$$
Therefore, $$0.625$$ moles of methanol are needed.
Molar mass of methanol is $$32g/mol$$. Hence, the mass of methanol for $$0.625$$ moles is,
Mass of methanol $$=0.625mol\times {\displaystyle \frac{32g}{mol}}=20g$$
Density of methanol $$=0.793kg/L$$ or $$793g/L$$
Hence, volume $$={\displaystyle \frac{mass}{density}}$$
Volume of methanol needed for $$20g$$ $$={\displaystyle \frac{20g}{793g/L}}=0.0252L=25.2mL$$
Hence, $$25.2mL$$ of methanol is required for making $$2.5L$$ of its $$0.25M$$ solution.

Additional Information:
Molarity is a convenient method in usage. This is because the molarity can be easily measured in grams. They can be converted into moles. Also, they can be mixed with volume. Molarity is an intensive property. Intensive properties are those that do not depend on the quantity or size of the matter existing in the system. Irrespective of the quantity of solution considered, whether large or small, the morality of the solution will be the same.

Note:
The total molar concentration can be easily obtained by adding up all the molar concentrations. Such a summation may be permitting the calculations of density and the strength of ions. However, one big disadvantage of molarity is that the molarity changes along with the change in temperature. This is due to the reason that the volume of a liquid is directly affected by the temperature. However, this problem can be solved if there exists no problem if the relevant measurements are conducted at a single temperature.