When 1 liter of air is burnt with a mixture calcium carbide and anhydrous calcium chloride, the reduction in volume of air is about

Air generally contains ${\mathrm{N}}_2,{\mathrm{O}}_2$ and ${\mathrm{H}}_2\mathrm{O}$. When this $1\mathrm{l}\left(1000\mathrm{ml}\right)$ of air is burnt with ${\mathrm{CaC}}_2$ gives ${\mathrm{CaCN}}_2$. And when this air burnt with a mixture of anhydrous ${\mathrm{CaCl}}_2$, the water is absorbed here. In $1000\mathrm{ml}$ of air, the presence of noble gases is $1\%$.

So, $1\%$ of $1000\mathrm{ml}=\left(\frac{1}{100}\right)\times 1000=10\mathrm{ml}$

So, the volume of air present is $10\mathrm{ml}$. Hence, the reduction in the volume of air is $=1000\mathrm{ml}-$ $10\mathrm{ml}=990\mathrm{ml}$.

Therefore, option (B) is correct.