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**Hint:**Analyze the terms given in the question carefully and write a chemical reaction equation for the given reaction and balance the reaction equation. The volume of acetylene is given and one can relate this term with oxygen and find out the value of the volume of air which contains only $20\% $ oxygen.

**Complete answer:**

1) First of all let's understand the values given in the question and write a general chemical equation for the reaction as below,

${C_2}{H_2} + {O_2} \to C{O_2} + {H_2}O$

Now let’s balance this reaction equation as below,

$2{C_2}{H_2} + 5{O_2} \to 4C{O_2} + 2{H_2}O$

2) From the above-balanced reaction equation, we can say that to burn two moles of acetylene we need five moles of oxygen. We can write this in ration form as,

${\text{2 }}{{\text{C}}_2}{{\text{H}}_2}{\text{: 5 }}{{\text{O}}_2}$

${\text{2 : 5}}$

Now we need the volume of oxygen which is required and the volume of acetylene which is burned is ${\text{50 c}}{{\text{m}}^3}$ and we can put this value and relate as below,

${\text{50 : x}}$

Where the value ${\text{x}}$ is the volume of oxygen.

SO, the volume of oxygen required $ = 50 \times \dfrac{5}{2} = 125c{m^3}$

Hence, the volume of $125c{m^3}$ oxygen is required to burn ${\text{50 c}}{{\text{m}}^3}$ acetylene.

3) But air contains $20\% $ oxygen which is given in question we need to calculate the value for the ${\text{100\% }}$ oxygen required. We can calculate this value as below,

${\text{Pure volume of oxygen required = }}\dfrac{{125 \times 100}}{{20}} = 625c{m^3}$

4) Therefore, to burn ${\text{50 c}}{{\text{m}}^3}$ acetylene we need $625c{m^3}$ pure oxygen which shows option D as the correct choice of answer.

**Therefore, the correct option is D.**

**Note:**

While writing the reaction equation, making it a balanced equation is a very important step where one can get the values of the number of moles of each component and can relate with one another to find out the volume composition of each component in the given reaction.

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