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  4. Volume at NTP of 0.22 g of CO2 is same as that of

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Q. Volume at NTP of 0.22 g of $ C{{O}_{2}} $ is same as that of

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Solution:

$ Moles=\frac{weight}{molecular\,weight} $ $ =\frac{volume\,(in\,L.)}{22.4} $ $ \therefore $ $ \frac{weight}{molecular\,weight}=\frac{molecular\,weight}{weight\times 22.4} $ or $ \frac{1}{Volume\,(in\,L)}=\frac{molecular\,weight}{weight\times 22.4} $ For $ 0.22g\,C{{O}_{2}}, $ $ \frac{1}{Volume}=\frac{44}{0.22\times 22.4}=\frac{200}{22.4} $ For 0.01 g of hydrogen, $ \frac{1}{Volume}=\frac{2}{0.01\times 22.4}=\frac{200}{22.4} $ For $ 0.085g\,\,N{{H}_{3}} $ , $ \frac{1}{Volume}=\frac{17}{0.085\times 22.4} $ $ =\frac{200}{22.4} $ For 0.32 g gaseous $ S{{O}_{2}} $ , $ \frac{1}{Volume}=\frac{64}{0.32\times 22.4} $ $ =\frac{200}{22.4} $ Thus, all have the same volume at NTP.