A) 3.44 ppm
B) 24 ppm
C) 34.4 ppm
D) 2.44 ppm
Correct Answer: B
Solution :
10 mL water\[=1.47\times {{10}^{-3}}g\,\,\]of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] \[{{10}^{6}}\]mL water \[=147\,g\]of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] \[\because \] 49 g of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\]requires \[{{O}_{2}}=89\] \[\therefore \] 147 g of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] will required \[{{O}_{2}}=\frac{8\times 147}{49}\] \[=24\,g\]
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