Question

Molarity of liquid \[HCl\] will be, if density of solution is \[1.17gm/cc\]?

Answer 1

Hint: Molarity represents the strength of a substance in solution. It is the ratio of the moles of solute dissolved in a litre of solution.

Complete step by step answer: \[HCl\] is hydrochloric acid whose density is \[1.17gm/cc\]. We know that \[1cc\] is equal to \[1mL\]as \[cc\] and \[mL\] are different names of the same volume. The unit \[cc\] stands for cubic centimeters and \[mL\] stands for milliliters.
\[1mL\] of \[HCl\] contains \[1.17gm\]. Therefore \[1L\] or \[1000mL\] of \[HCl\] contains \[1.17 \times 1000\]= \[1170gm\].
In order to determine the moles of \[HCl\] present in \[1170gm\] we need to determine the molar or molecular mass of \[HCl\].
Molar mass of \[HCl\] = atomic mass of \[H\] + atomic mass of \[Cl\].
$\Rightarrow$ \[1 + 35.5\] = \[36.5gm\].
Thus the moles of \[HCl\] in \[1170gm\] = $\dfrac{{weight{\text{ }}of{\text{ }}HCl}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}HCl}}$
$\Rightarrow$ $\dfrac{{1170}}{{36.5}}$ = \[32.05moles\].
As, morality is determined as the ratio of number of moles and volume of solution in litre, so the molarity of \[HCl\] liquid will be
Molarity = $\dfrac{{no.{\text{ }}of{\text{ }}moles}}{{volume{\text{ }}in{\text{ }}litre}}$
$\Rightarrow$ $\dfrac{{32.05}}{1}$ = \[32.05M\].

Note: Molarity is also referred to as molar concentration of a solution. Similar term molality should not be confused with morality. Molarity is expressed in upper case as \[M\] and molality is expressed in lower case as \[m\]. The basic difference between the two lies in mass vs. volume. Molality is the ratio of moles of solute dissolved in kilograms of solvent. So the unit is \[mol/kg\] or \[m\]. Both are the measure of a substance concentration in solution. Density is the measure of compactness of a substance in solid, liquid or gas phase. It is the ratio of mass and volume of the substance. So more the density of a substance less will be the volume and vice versa. The density of water is \[1gm/mL\].