Question

The molal freezing point depression constant for benzene ${C_6}{H_6}$ is $4.90KKgmo{l^{ - 1}}$. Selenium exists as a polymer of the type $S{e_x}$, of benzene, the observed freezing point is $0.112^\circ C$ lower than that of pure benzene. The molecular formula of selenium is:
(atomic mass of Se = $78.8gmo{l^{ - 1}}$)
(A) $S{e_8}$
(B) $S{e_6}$
(C) $S{e_4}$
(D) $S{e_2}$

Answer 1

Further the molal freezing point depression constant is basically the constant that is equal to the change in the freezing point for a 1 molal solution of a nonvolatile molecular solute. the molar mass of selenium which can be further calculated by the formula:

Formula used: ${M_B} = \frac{{{K_f} \times {W_B}}}{{\Delta {T_f} \times {W_A}}}$
This formula will further help to find the value of the x for the selenium.

Complete step-by-step solution: Firstly we will discuss the freezing point depression so it is basically the difference in the freezing point of the pure solvent and that of the solution. This happens because when the solute is added in the pure solvent at that time the vapour pressure of the solvent resulting in the lowering of the freezing point of the solution compared to the solvent. Further the molal freezing point depression constant is basically the constant that is equal to the change in the freezing point for a 1 molal solution of a nonvolatile molecular solute.
Now coming to the question we need to find the value of x in the polymer of the selenium. So we are given with :
atomic mass of Se = $78.8gmo{l^{ - 1}}$
the molal freezing point depression constant for benzene ${C_6}{H_6}$ is $4.90KKgmo{l^{ - 1}}$
the observed freezing point is $0.112^\circ C$
then firstly we need to calculate the molar mass of selenium which can be further calculated by the formula:
${M_B} = \frac{{{K_f} \times {W_B}}}{{\Delta {T_f} \times {W_A}}}$
Where ${K_f}$is the molal freezing point depression constant, ${W_B}$is the mass of selenium, $\Delta {T_f}$is the freezing depression observed, ${W_A}$is the mass of benzene.
Now putting the values in the equation we get:
$\frac{{4.9 \times 3.26}}{{0.112 \times 0.226}} = 631.08gmo{l^{ - 1}}$
Moreover it is given that the atomic mass of Se = $78.8gmo{l^{ - 1}}$
So further the value of x can be calculated as:
$\frac{{631.08}}{{78.8}} = 8$

Hence the correct option is (A).

Note: From the above calculation we concluded that the correct option is option A and to find the correct value of x we need to carry the calculations with much more accuracy. Each and every step should be calculated by putting the accurate formula.