A) 0.41
B) 0.35
C) 0.48
D) 0.54
Correct Answer: C
Solution :
The molar mass of \[{{\text{O}}_{2}}=32g/mol\] \[0.16g\] of oxygen \[=\frac{0.16g}{32g/mol}=0.005mol\] \[2NaCl{{O}_{3}}=2NaCl+3{{O}_{2}}\] 3 moles of \[{{O}_{2}}=2\] moles of \[NaCl=2\] moles of \[AgCl.\] 0.005 moles of \[{{O}_{2}}=0.005\times \frac{2}{3}=0.003333\] moles of \[AgCl\]. Molar mass of \[AgCl=143.5g\,\,mo{{l}^{-1}}\] The mass of \[AgCl\] (in g) obtained will be \[=143.5g\,\,mo{{l}^{-1}}\times 0.003333mol=0.48g.\]You need to login to perform this action.
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