Answer:
(i) Volume of 1.6 g\[{{O}_{2}}\] at STP
\[=\frac{1.6}{32}\times
22.4=1.12L\]
\[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\]
or \[1\times
1.12=0.5\times {{V}_{2}}\]
or \[{{V}_{2}}=2.24L\]
(ii) Number of molecules =
\[\frac{Mass}{molar\,mass}\times
6.023\times {{10}^{23}}\]
\[=\frac{1.6}{32}\times
6.023\times {{10}^{23}}\]
\[\mathbf{=3}\mathbf{.0115\times
1}{{\mathbf{0}}^{\mathbf{22}}}\]
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