Question

An aqueous solution of $FeC{l_3}$ (molar mass $= 162gmo{l^{ - 1}}$ has a density of 1.1 g/mL and contains $20\%$ $FeC{l_3}$ . What is the molarity of this solution?
A)0.28 M
B)1.35 M
C)2.17 M.
D)1.97 M

Answer 1

Molarity is used to calculate the volume of a solvent in a solution. Molarity is inversely proportional to volume i.e. volume increases if the molarity of the solution decreases.
Formula used: $m = \dfrac{{moles}}{{volume(in\,litres)}} = \dfrac{{w \times 1000}}{{M \times V}}$

Complete step by step answer:
In general, molarity is referred to the total number of moles of solute per liter volume of solution. It depends on the changes in pressure, volume, and temperature. Its symbol M stands for molar unit concentration, which means molarity of a solution when one gram of solute is dissolved in one liter of solution.
It can be calculated by the following formula
$m = \dfrac{{moles}}{{volume(in\,litres)}} = \dfrac{{w \times 1000}}{{M \times V}}$
Where, w is the mass of the solute, M is molecular mass, V is the volume of the solution.
From this formula, we derive another formula:
${M_1}{V_1} = {M_2}{V_2}$
This equation is used when a stock solution of higher polarity is provided and we are supposed to dilute it to lower molarity by adding more solvent. This dilution formula is used to calculate any of the four quantity when values of the rest of the three are provided.
Now according to the question, let the total mass of the solution is 100gm. Now, it means $20\%$ $FeC{l_3}$ is equal to 20 gm $FeC{l_3}$ .
Now, density, $d = \dfrac{w}{M}$ , where, w is the mass of solvent and V is the volume.
Now, the volume of the solvent is,

$$d = \dfrac{w}{V} \\\ V = \dfrac{w}{d} \\\ V = \dfrac{{100}}{{1.1}} \\\ V = \dfrac{{100}}{{1.1}} \\\$$

Now the molality is,

$$m = \dfrac{{moles}}{{volume(in\,litres)}} = \dfrac{{w \times 1000}}{{M \times V}} \\\ m = \dfrac{{20 \times 1000}}{{162 \times \dfrac{{100}}{{1.1}}}} \\\ m = \dfrac{{20 \times 1000 \times 1.1}}{{162 \times 100}} \\\ m = \dfrac{{20 \times 10 \times 1.1}}{{162}} \\\ m = \dfrac{{20 \times 11}}{{162}} \\\ m = 1.35M \\\$$

So, the correct answer is B.

Note: Molality and molarity are two different concepts. Students usually get confused between these two. Hence,
Molality is the ratio of the number of moles of the solute to the total mass of the solvent. Representing this in the form of an equation:
Molality $= \dfrac{{no.\,of\,moles\,of\,solute}}{{mass\,of\,solvent}}$