A gaseous mixture contains CH 4 and C 2 H 6 in equimolecular proportion. The weight of 2.24 litres of this mixture at NTP is

# A gaseous mixture contains ${\mathrm{CH}}_{4}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{C}}_{2}{\mathrm{H}}_{6}$ in equimolecular proportion. The weight of 2.24 litres of this mixture at NTP is

1. A
4.6 g
2. B
2.3 g
3. C
1.6 g
4. D
23 g

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### Solution:

Equimolecular proportion means both gases occupied equal volume

Volume of the mixture = 2.24l

volume of   ${\mathrm{CH}}_{4}=\frac{2.24}{2}\text{\hspace{0.17em}}=\text{\hspace{0.17em}\hspace{0.17em}}1.12\text{\hspace{0.17em}}\mathrm{L}$

$\begin{array}{l}22.4\text{\hspace{0.17em}}\mathrm{L}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{CH}}_{4}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{has}\text{\hspace{0.17em}}\mathrm{mass}=16\text{\hspace{0.17em}}\mathrm{g}\\ 1.12\text{\hspace{0.17em}}\mathrm{L}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{CH}}_{4}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{has}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{mass}=\frac{16}{22.4}\text{\hspace{0.17em}}×\text{\hspace{0.17em}\hspace{0.17em}}1.12=0.8\text{\hspace{0.17em}}\mathrm{g}\\ \mathrm{For}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{C}}_{2}{\mathrm{H}}_{6}\\ 22.4\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{L}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{C}}_{2}{\mathrm{H}}_{6}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{has}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{mass}\text{\hspace{0.17em}}=\text{\hspace{0.17em}\hspace{0.17em}}30\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{g}\\ 1.12\text{\hspace{0.17em}}\mathrm{L}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{C}}_{2}{\mathrm{H}}_{6}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{has}\text{\hspace{0.17em}}\mathrm{mass}\text{\hspace{0.17em}}\frac{30}{22.4}×\text{\hspace{0.17em}\hspace{0.17em}}1.12=\text{\hspace{0.17em}\hspace{0.17em}}\frac{3.0}{2}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}1.5\mathrm{g}\\ \mathrm{Total}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{mass}=1.5\text{\hspace{0.17em}}\mathrm{g}+0.8\mathrm{g}=2.3\text{\hspace{0.17em}}\mathrm{g}.\end{array}$

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