Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

If $60mL$ of a mixture of equal volume of $C{l_2}$ and an oxide of chlorine was heated and then cooled back to the original temperature. The resulting gas mixture was found to have a volume of $75mL$. On treatment with caustic soda solution, the volume contracted to $15mL$ . Assume that all measurements are made at the same $T$ and $P$ . Deduce the simplest formula for oxide of $C{l_2}$ and calculate its molar mass. The oxide of $C{l_2}$ on heating decomposes quantitatively to ${O_2}$ and $C{l_2}$ . If the molar mass is $x$ $g/mol$, then find the nearest integral value of $x$ .

seo-qna
Last updated date: 18th Jun 2024
Total views: 393.9k
Views today: 9.93k
Answer
VerifiedVerified
393.9k+ views
Hint: When the mixture is heated, the gases which will be left are chlorine and oxygen. We are given that this is then passed through caustic soda, which will absorb all the chlorine. As we are given the volume after this process, we can find moles of oxygen by comparing it to the initial process, and then find the total molecular mass.

Complete step by step answer:
After heating the mixture, chlorine gas will remain as it is while the oxide will decompose to chlorine and oxygen. As they are of equal volume in the beginning, the chlorine gas and the chlorine oxide will occupy $\dfrac{60mL}{2} = 30mL$ each. Let $n$ denote the number of atoms of oxygen in the oxide. Thus, we can write a general formula for the oxide as $C{l_2}{O_n}$ . We can write a general equation for the decomposition from the data we’re given as:

$C{l_2} + C{l_2}{O_n}\xrightarrow{\Delta }2C{l_2} + (n/2){O_2}$

Initial volumes ($mL$)$30$$30$$0$$0$
Final volumes ($mL$)$0$$0$$60$$15n$


We know that one molecule of chlorine occupies $30mL$. Therefore, after the heating, the two molecules of chlorine will occupy a combined volume of $60mL$. As the oxide is a gas, the volume occupied by the $n$ atoms of oxygen will also be $30mL$. Hence, after heating when this has been converted to molecular oxygen, $\dfrac{n}{2}$ moles will occupy a volume of $\dfrac{30n}{2} = 15n$.
Now, after passing through the caustic soda solution, all of the chlorine will be absorbed and we are left with only the oxygen gas. The volume of this is given as $15mL$ . Hence, equating the volume of oxygen we got from the heating to this volume, we get:
$15n = 15 \Rightarrow n = 1$
Substituting the value of $n$ in our general formula of the oxide, we get the oxide to be $C{l_2}O$.
Hence, its molecular mass is:
$(35.5 \times 2) + 16$ since molecular mass of $Cl = 35.5$ and $O = 32$.
Therefore, molar mass of the compound $ = 71 + 16 = 87g/mol$

Thus, the nearest integral value of $x = 87$

Note: Chlorine can form many oxides ranging from acidic to basic, and most of these are unstable and release free chlorine. Hence, they are used as good chlorinating and bleaching agents. Note that the chlorine gas reacts with caustic soda ($NaOH$) to produce sodium chlorate and sodium chloride, both solids, and water. That is why oxygen is the only gaseous compound present after passing the mixture through caustic soda.