Question

Mass of solid ${I_2}$ (MW $ = $ 254) required to dissolve in 0.5 kg $CC{l_4}$ to prepare 0.1 m solution is
A. 50.8 g
B. 25.4 g
C. 12.7 g
D. 17.8 g

Answer 1

Hint: Composition of a solution can be expressed on the basis of 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:
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

$Molality(m) = \dfrac{{{{Moles of solute}}}}{{{{Mass of solvent in kg}}}}$

Moles of the solute can be calculated by the formula

${{Moles = }}\dfrac{{Mass}}{{Molar\,Mass}}$

In the above question we can identify the solute as ${I_2}$ with the given molar mass $ = $ 254. Using the above formula we can calculate the moles of the solute as

Moles of solute = $\dfrac{{Mass\,of\,solute}}{{Molar\,Mass\,of\,solute}}$

$\Rightarrow Moles\,of\,solute = \dfrac{{Mass\,of\,solute}}{{254}}$

Mass of the solvent $CC{l_4}$ in kg is given in the question as 0.5 kg while the molality of the solution is given as 0.1 m
Putting the above information in the equation of molality we get

Molality = $\dfrac{{Moles\,of\,solute}}{{Mass\,of\,solvent\,in\,kg}}$

$\Rightarrow 0.1 = \dfrac{{\dfrac{{Mass\,of\,solute}}{{254}}}}{{0.5}}$
After cross multiplication we get

$\Rightarrow 0.05 = \dfrac{{Mass\,of\,solute}}{{254}}$

$\Rightarrow$ 12.7 = Mass of ,solute
Therefore the mass of solute is 12.7 g

So, the correct answer is Option C.

Note: There are other representations of the concentration of solution than molality.
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 in the case of molality. It can be represented as
${{Molarity (M) = }}\dfrac{{{{\text{Moles of solute}}}}}{{{{\text{Volume of solution in litre}}}}}$