Ans :- 0.16 M
Explanation :-
Number of moles = Given mass in g / Gram molar mass
Given mass of Cu(NO3)2 = 3.03 g
Gram molar mass of Cu(NO3)2 = 187.56 g/mol
So,
Number of moles of Cu(NO3)2 = 3.03 g / 187.56 g/mol
= 0.01615 mol
Given Volume of Cu(NO3)2 solution = 200 mL = 0.200 L
Molarity = Number of moles of solution / Volume of solution in L
So,
Molarity of Cu(NO3)2 solution = 0.01615 mol / 0.200 L
= 0.08075 M
Given, Molarity of Na2CrO4 = 72.0 mM = 0.072 M
Now,
ICF table is :
.........................Cu(NO3)2 (aq)..............+.............Na2CrO4 (aq) <----------------> CuCrO4 (aq)..........+.........2 NaNO3 (aq)
Initial (I)................0.08075 M.................................0.072 M..................................0.0 M.....................................0.0 M
Change (C)..........-0.072 M....................................-0.072 M.................................+0.072 M..............2x0.072 M = 0.144 M
Final (F)................0.00875 M....................................0.0 M...................................0.072 M................................0.144 M
In Cu(NO3)2 (aq) a excessive reagent , Concentration of NO3- (M1) = 2 x 0.00875 M = 0.0175 M
In NaNO3 (aq) , Concentration of NO3- (M2) = 0.144 M
So,
Final or Total concentration of NO3- in the solution = M1 + M2 = 0.0175 M + 0.144 M = 0.1615 M
Hence, Final concentration of NO3- in solution = 0.16 M |
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