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  4. The number of gram-molecules of oxygen which are present in 6.022× 1024CO molecules is

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Q. The number of gram-molecules of oxygen which are present in $ 6.022\times {{10}^{24}}CO $ molecules is

JamiaJamia 2011

Solution:

Number of oxygen atoms = Number of CO molecules $ =6.022\times {{10}^{24}} $ Number of oxygen molecules $ =\frac{1}{2}\times $ No. of oxygen atoms $ =\frac{1}{2}\times 6.022\times {{10}^{24}}=3.011\times {{10}^{24}} $ Number of g-molecules of oxygen $ =\frac{3.011\times {{10}^{24}}}{6.022\times {{10}^{23}}}= $ 5 g-molecule