Comprehension-1 Eudiometery is the technique to determine the molecular formula of gases as well...
Comprehension-1
Eudiometery is the technique to determine the molecular formula of gases as well as composition of gases in a gaseous mixture by studying the volumes of gases at same pressure and temperature. During the analysis, the eudiometer tube filled with mercury is inverted over a trough containing mercury. A known volume of the gas or gaseous mixture to be studied is next introduced, which displaces an equivalent amount of mercury. Next a known amount of oxygen in excess is introduced and the electric spark is passed or explosion is made, whereby the combustible material gets oxidized. The volume of carbon dioxide, water vapour or other gaseous products of combustion are then determined by absorbing them in suitable reagents. The various reagents used for absorbing different gases are:
\( \begin{aligned} \mathrm{O}_{3} & \longrightarrow \text { turpentine oil } \\ \mathrm{CO}_{2}, \mathrm{SO}_{2} \text { and all other gaseous oxides } & \longrightarrow \text { alkaline pyrogallol } \\ \mathrm{NO} & \text { alkali solution }\left[\mathrm{NaOH}, \mathrm{KOH}, \mathrm{Ca}(\mathrm{OH})_{2}\right] \\ \mathrm{O}_{2} \text { and } \mathrm{N}_{2} & \longrightarrow \mathrm{FeSO}_{4}(\mathrm{aq}) \\ \mathrm{H}_{2} \mathrm{O}_{2} \text { and } \mathrm{NH}_{3} \longrightarrow \text { heated } \mathrm{Mg} \\ \mathrm{CO} \text { and } \mathrm{C}_{2} \mathrm{H}_{2} & \longrightarrow \text { conc. } \mathrm{H}_{2} \mathrm{SO}_{4} \\ \end{aligned} \)
A gas mixture of 3 litre of propane and butane on complete combustion at \( 25^{\circ} \) produced 10 litre of \( \mathrm{CO}_{2} \). The gas mixture contains initially:
(A) 2 litre \( \mathrm{C}_{3} \mathrm{H}_{8}+1 \) litre \( \mathrm{C}_{4} \mathrm{H}_{10} \)
(B) 1 litre \( \mathrm{C}_{3} \mathrm{H}_{8}+2 \) litre \( \mathrm{C}_{4} \mathrm{H}_{10} \)
(C) \( 1.5 \) litre \( \mathrm{C}_{3} \mathrm{H}_{8}+1.5 \) litre \( \mathrm{C}_{4} \mathrm{H}_{10} \)
(D) \( 2.5 \) litre \( \mathrm{C}_{3} \mathrm{H}_{8}+0.5 \) litre \( \mathrm{C}_{4} \mathrm{H}_{10} \)
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