A) \[0.1\times {{10}^{-3}}\]
B) \[0.5\times {{10}^{-3}}\]
C) \[1.66\times {{10}^{-3}}\]
D) \[9.95\times {{10}^{-2}}\]
Correct Answer: B
Solution :
[b] Moles of \[{{H}_{2}}S{{O}_{4}}\]in 98 mg of \[{{H}_{2}}S{{O}_{4}}\]\[=\frac{1}{98}\times 0.098=0.001\] |
Moles of \[{{H}_{2}}S{{O}_{4}}\]removes \[=\frac{3.01\times {{10}^{20}}}{6.02\times {{10}^{23}}}=0.5\times {{10}^{-3}}=0.0005\] |
Moles of \[{{H}_{2}}S{{O}_{4}}\]left \[=0.001-0.0005=0.5\times {{10}^{-3}}\] |
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