What is the mass of the precipitate formed when $$50$$ mL of $$16.9$$ % solution of $$AgNO3$$ is mixed with $$50$$ mL of $$5.8$$% NaCl solution? (Ag=107.8$$, $$N=14$$,$$O=16$$,$$Na=23$$,$$Cl=35.5)

Question

What is the mass of the precipitate formed when $$50$$ mL of $$16.9$$ %  solution of $$AgNO3$$ is mixed with $$50$$ mL of $$5.8$$% NaCl solution? (Ag=107.8$$, $$N=14$$,$$O=16$$,$$Na=23$$,$$Cl=35.5)

Infinity Learn · Accepted Answer

$$16.9$$% solution of $$AgNO3$$ means $$16.9$$ g of $$AgNO3$$ in $$100$$ mL of $$solution.16.9$$ g of $$AgNO3$$ in $$100$$ mL solution $$=$$ $$8.45$$ g of $$AgNO3$$ in $$50$$ mL solution. Similarly, $$5.8$$% of NaCl in $$100$$ mL solution $$=$$ $$2.9$$ g of NaCl in $$50$$ mL solution.The reaction can be represented as :                      $$AgNO3+NaCl$$⟶$$AgCl+NaNO3Initially$$       $$8.45/170$$        $$2.9/58.5$$                $$=$$  $$0.049$$ mol    $$=$$ $$0.049$$ molFinally            $$0$$                        $$0$$           $$0.049$$ mol    $$0.049$$ mol∴  Mass of AgCl precipitated $$=0.049$$×$$143.3$$                                                                         $$=7.02$$≈$$7$$ g

What is the mass of the precipitate formed when 50 mL of 16.9 % solution of AgNO3 is mixed with 50 mL of 5.8% NaCl solution? (Ag=107.8, N=14,O=16,Na=23,Cl=35.5)

Detailed Solution

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