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`1 L` of `0.1 M NaOH, 1 L` of `0.2 M KOH`, and `2 L` of `0.05 M Ba (OH)_(2)` are mixed togther. What is the final concentration of the solution.A. 0.01 MB. 0.01 NC. 0.1 ND. 0.001 M |
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Answer» Total volume `(V_4)=4L` n factor for NaOH and KOH is 1 whereas that for `Ba(OH)_2` is 2 Now, `N_1V_1+N_2V_2+N_3V_3=N_4V_4` `0.1xx1+0.2xx1+0.05xx2xx2=N_4xx4` `rArr 0.1 + 0.2 + 0.2 = N_4xx4 rArr N_4 = 0.125` Hence, final concentration of solution = 0.1 N |
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