Question

If $36.5\% {\text{ HCl}}$ has density equal to $1.20{\text{ g m}}{{\text{l}}^{ - 1}}$. The molarity (M) and molality (m) respectively are:
A) 15.7, 15.7
B) 12, 12
C) 15.7, 12
D) 12, 15.7

Answer 1

To solve this we must know the terms molarity and molality. Both molarity and molality are used to express the concentration of solution. Remember that to calculate the molality we need the mass of the solvent and not the solution.

Complete solution:
We know that density is the ratio of mass to the volume. The expression for density is as follows:
$d = \dfrac{m}{V}$
Where $d$ is the density of the solution,
$m$ is the mass of the solution,
$V$ is the volume of the solution,
Rearrange the equation for the mass of the solution as follows:
$m = d \times V$
Substitute $1.20{\text{ g m}}{{\text{l}}^{ - 1}}$ for the density, $1{\text{ L}} = 1000{\text{ mL}}$ for the volume of the solution. Thus,
$m = 1.20{\text{ g m}}{{\text{l}}^{ - 1}} \times 1000{\text{ mL}}$
$m = 1200{\text{ g}}$
Thus, the mass of the solution is $1200{\text{ g}}$.
We are given $36.5\% {\text{ HCl}}$. Thus,
${\text{Mass of HCl}} = 1200{\text{ g}} \times \dfrac{{36.5}}{{100}} = 438{\text{ g}}$
We know that the number of moles is the ratio of mass to the molar mass. Molar mass of ${\text{HCl}}$ is $36.5{\text{ g mo}}{{\text{l}}^{ - 1}}$. Thus,
${\text{Number of moles of HCl}} = \dfrac{{438{\text{ g}}}}{{36.5{\text{ g mo}}{{\text{l}}^{ - 1}}}} = 12{\text{ mol}}$
We know that molarity is the number of moles of solute per litre of solution.
Thus, the molarity of $36.5\% {\text{ HCl}}$ is $12{\text{ mol}}$ in one litre of solution i.e. $12{\text{ M}}$.
Now, we know that the mass of the solution is $1200{\text{ g}}$ and the mass of the solute is $438{\text{ g}}$. Thus,
${\text{Mass of solvent}} = 1200 - 438 = 762{\text{ g}}$
Thus, $12{\text{ mol}}$ of solute is present in $762{\text{ g}}$ of solvent. Thus, $1000{\text{ g}}$ of solvent contains,
${\text{Number of moles of HCl}} = \dfrac{{12{\text{ mol}}}}{{762{\text{ g}}}} \times 1000{\text{ g}} = 15.7{\text{ mol}}$
We know that molality is the number of moles of solute per kilogram of solvent.
Thus, the molarity of $36.5\% {\text{ HCl}}$ is $15.7{\text{ mol}}$ in one kilogram of solvent i.e. $15.7{\text{ m}}$.
Thus, molarity (M) and molality (m) respectively are 12 and 15.7.

Thus, the correct option is (D) 12, 15.7.

Note: Remember that molarity is the number of moles of solute per litre of solution. And molality is the number of moles of solute per kilogram of solvent. The unit of molarity is ${\text{mol }}{{\text{L}}^{ - 1}}$ or ${\text{M}}$. And the unit of molality is ${\text{mol k}}{{\text{g}}^{ - 1}}$ or ${\text{m}}$.