A) 44.8 L
B) 11.2 L
C) 22.4 L
D) 5.6 L
Correct Answer: D
Solution :
\[\frac{{{W}_{1}}}{{{E}_{1}}}=\frac{{{W}_{2}}}{{{E}_{2}}}\] \[\frac{4.5}{9}\,\,=\,\,\frac{{{W}_{2}}}{1}\] \[{{W}_{2}}\,\,=\,\,0.5\text{ }g~~~\,\,\,\,\therefore \,\,\text{ }moles\,\,=\,\,\frac{0.5}{2}\] \[and\text{ }vol.=\frac{0.5}{2}\,\,\times \,\,22.4\,\,\,=\,\,\,5.6\,\,l\]You need to login to perform this action.
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