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How many moles of `BaCl_(2)` would be needed to make 250 ml of a solution having the same concentration of `Cl^(-)` as one containing 1.17 g NaCl per 100ml?A. 0.1B. 0.01C. 0.05D. 0.025 |
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Answer» Correct Answer - D `[Cl^(-)]_(NaCl)=(1.17)/(58.5)xx(1000)/(100)=[Cl^(-)]_(BaCl_(2))` `therefore [Cl^(-)]_(BaCl_(2))=("mole "Cl^(-))/((250)/(1000))=0.2` Mole of `Cl^(-)=0.2 xx(1)/(4)` `therefore` mole of `BaCl_(2) = 0.2 xx(1)/(4)xx(1)/(2)=0.025` |
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