Question

The electrolytic composition of water gives $H$ and ${O_2}$ in the ratio of
$(1)$ $1:2$ by volume
$(2)$ $2:1$ by volume
$(3)$ $8:1$ by mass
$(4)$ $1:2$ by mass

Answer 1

Electrolysis is the process of passing an electric current through a liquid containing ions so that decomposition reaction takes place. For comparing volumes, assuming both gases to be ideal.

Complete answer:
The formula of water is ${H_2}O$, water molecules are made up of one ${H_2}$ molecule and half oxygen molecule ${H_2} + \dfrac{1}{2}{O_2} \to {H_2}O$. So when the process is reversed by electrolysis, By the law of mass action we can say that ${H_2}O \to {H_2} + \dfrac{1}{2}{O_2}$. This means that in every molecule of water there are $2$ atoms of hydrogen and $1$ atoms of oxygen. So if $1$ molecule of water is electrolyzed then $2$ atoms of hydrogen and $1$ atoms of oxygen is produced.
During the electrolysis of water a redox reaction can be watched. At the cathode a reduction happens and on the anode oxidation. Normally, the reaction is balanced with either acid or base as shown in the equation below.
The reduction at cathode $:2{H^ + }(g) + 2{e^ - } \to {H_2}(l)$
The oxidation at anode $:2{H_2}O(g) \to {O_2}(g) + 4{e^ - } + 4{H^ + }(l)$
So, When an electric current is passed through acidified water (normally hydrochloric acid or sulfuric acid), it decomposes to give hydrogen and oxygen gas. The hydrogen gas is obtained at the cathode (negative electrode) and balanced with a base $4O{H^ - }(l) \to {O_2}(g) + 2{H_2}O(l) + 4{e^ - }$ (sodium hydroxide) $2{H_2}O(l) + 2{e^ - } \to {H_2}(g) + 2O{H^ - }(l)$
And the oxygen gas is obtained at the anode (positive electrode) $4O{H^ - }(l) \to {O_2}(g) + 2{H_2}O(l) + 4{e^ - }$
Combining the pair of the reactions, lead to an overall reaction which is the decomposition of water into oxygen and hydrogen are as follow $:2{H_2}O(l) \to 2{H_2}(g) + {O_2}(g)$
Thus the number of hydrogen molecules produced is twice the number of oxygen molecules. So, the electrolysis of water produces $2$ volumes of hydrogen and $1$ volume of oxygen gas because the ratio of hydrogen and oxygen elements in water is $2:1$ by volume. Therefore, the amount of gases collected in two test tubes is not equal.
And hence option 2 is the correct answer.

Note:
Look at the stoichiometric equation that represents the reaction. ${H_2}O(l)\xrightarrow{\Delta }{H_2}(g) + \dfrac{1}{2}{O_2}(g)$
Now, as gas under equal conditions of temperature and pressure, which surely must apply here, the Volume expressed by each gas represents their molar proportion. And thus dihydrogen expresses twice the volume of dioxygen.
And in fact, because the dihydrogen is the product of reduction, and dioxygen the product of oxidation, the gases could be collected at the appropriate terminus of the electrolytic cell. On the other hand the masses of the gases collected are respectively $2$g$mo{l^{ - 1}}$ of dihydrogen, but $32$ g $mo{l^{ - 1}}$ of dioxygen.