Question

Given, $3{C_2}{H_2}\xrightarrow{{red\;hot\;tubes}}X$
$2N{a_2}{S_2}{O_3} + Y \to N{a_2}{S_4}{O_6} + 2NaI$
Mole fraction of Y in X is 0.2. What is the molality of Y in X?

Answer 1

We will first need to know what the values of X in the given equation is. Similarly we will need the value of Y too. After knowing what X and Y stand for we can directly find the molality of Y in X by using the molality formula.
Formula used:
$m = \dfrac{{{n_y}}}{{{n_x} \times \dfrac{{{M_x}}}{{1000}}}}$
Where, $m$= molality
${n_y}$=number of moles of Y
${n_x}$=number of moles of X
${M_x}$=molar mass of X

Complete answer:
We know that according to the equation,
$3{C_2}{H_2}\xrightarrow{{red\;hot\;tubes}}X$
X will be ${C_6}{H_6}$
And by looking at the equation
$2N{a_2}{S_2}{O_3} + Y \to N{a_2}{S_4}{O_6} + 2NaI$
We can easily say that Y is ${I_2}$
As the question relates mole fraction and molality we should know that molar mass of benzene is 78.
The molar mass of iodine gas is 254.
The aqueous 10 moles of solution has 8 moles of benzene and 2 moles of iodine.
Hence molality will be:
$m = \dfrac{2}{{8 \times \dfrac{{78}}{{1000}}}}$
Or, $m = 3.2$

Hence the correct answer to this question is option a.

Additional information:
When three molecules of ethyne are sent through a red hot tube, they cyclically polymerize and generate an aromatic compound with six carbon atoms. When ethyne is pushed through a red hot iron tube at 873 K, it polymerizes. Polymerization is the chemical reaction that joins monomers or smaller compounds to form three-dimensional linkages and polymer chains.
Three ethyne molecules cyclically polymerize to form benzene in this case.

Note:
Remember the equations correctly, here X stands for benzene and not ethylene. Similarly Y has to be iodine. And the number of moles of both X and Y should also be correct.