Solveeit Logo

Question

Question: If \(60mL\) of a mixture of equal volume of \(C{l_2}\) and an oxide of chlorine was heated and then ...

If 60mL60mL of a mixture of equal volume of Cl2C{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 75mL75mL. On treatment with caustic soda solution, the volume contracted to 15mL15mL . Assume that all measurements are made at the same TT and PP . Deduce the simplest formula for oxide of Cl2C{l_2} and calculate its molar mass. The oxide of Cl2C{l_2} on heating decomposes quantitatively to O2{O_2} and Cl2C{l_2} . If the molar mass is xx g/molg/mol, then find the nearest integral value of xx .

Explanation

Solution

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 60mL2=30mL\dfrac{60mL}{2} = 30mL each. Let nn denote the number of atoms of oxygen in the oxide. Thus, we can write a general formula for the oxide as Cl2OnC{l_2}{O_n} . We can write a general equation for the decomposition from the data we’re given as:

Cl2+Cl2OnΔ2Cl2+(n/2)O2C{l_2} + C{l_2}{O_n}\xrightarrow{\Delta }2C{l_2} + (n/2){O_2}

Initial volumes (mLmL)303030300000
Final volumes (mLmL)0000606015n15n

We know that one molecule of chlorine occupies 30mL30mL. Therefore, after the heating, the two molecules of chlorine will occupy a combined volume of 60mL60mL. As the oxide is a gas, the volume occupied by the nn atoms of oxygen will also be 30mL30mL. Hence, after heating when this has been converted to molecular oxygen, n2\dfrac{n}{2} moles will occupy a volume of 30n2=15n\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 15mL15mL . Hence, equating the volume of oxygen we got from the heating to this volume, we get:
15n=15n=115n = 15 \Rightarrow n = 1
Substituting the value of nn in our general formula of the oxide, we get the oxide to be Cl2OC{l_2}O.
Hence, its molecular mass is:
(35.5×2)+16(35.5 \times 2) + 16 since molecular mass of Cl=35.5Cl = 35.5 and O=32O = 32.
Therefore, molar mass of the compound =71+16=87g/mol = 71 + 16 = 87g/mol

Thus, the nearest integral value of x=87x = 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 (NaOHNaOH) 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.