InterviewSolution
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InterviewSolution
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Normally: An aquous solution contains `4.202 g` of `HNO_(3)` in `600 mL` of solution. Calculate the normally of solution? Stragegy: Convert grams of `HNO_(3)` to moles of `HNO_(3)` to moles of `HNO_(3)` and then to equivalents of `HNO_(3)`. Finally , apply the defintion of normally. `(g HNO_(3))/(L) rarr (mol HNO_(3))/(L) rarr (eq HNO_(3))/(L) rarr (eq HNO_(3))/(L) = N HNO_(3)` |
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Answer» Step 1: Calculate the number of moles. mol `HNO_(3) = ("Grams" HNO_(3))/("Molar mass" HNO_(3)) = (4.202g)/(63.02g mol^(-1))` `= 0.0666 mol HNO_(3)` Step 2: Calculate the number of equivalnets of `HNO_(3)`. Number of equivalnents of `HNO_(3)` (No. eq `HNO_(3)`) = (Number of moles of acid) `xx` (Number of hydrogen ions furnished) `= (0.0666 mol HNO_(3))(1)` `= 0.0666 eq HNO_(3)` Step 3: Calcularte normally. Normally `(N) = ("No. eq" HNO_(3))?("Liners of solution")` `= (0.0666 eq HNO_(3))/(0.600 L)` `= 0.11 NHNO_(3)` Alterntively using single setup, `? (eq HNO_(3))/(L) = (4.202 g HN O_(3))/(0.600 L) xx (1 mol HNO_(3))/(63.02g HNO_(3))` `xx (1 eq HNO_(3))/(1 mol HNO_(3))` `= 0.0111 N NHO_(3)` |
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