Question

Suppose 3.03 g of copper(II) nitrate is dissolved in 200. mL of a 72.0 m M aqueous solution of sodium chromate. Calculate the final molarity of copper(II) cation in the solution. You can assume the volume of the solution doesn't change when the copper(II) nitrate is dissolved in it. Round your answer to 2 significant digits. Пм x 6 ?

Answer 1

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