Answer:
0.367 g
Explanation:
Precipitation of silver and Cl requires 1:1 mol ratio of silver and Cl each, so the amount of mol NaCl (provides 1 mol of Cl⁻) and AgNO₃ (privides 1 mol of Ag⁺) must be equal for complete precipitation.
mol of Ag⁺ we have is
[tex]mol_{Ag^+}=[AgNO_3]*V_{solution}\\ =0.251*0.025=0.006275\ mol[/tex]
Thus we need 0.00625 mol of NaCl as well
The molar mass of NaCl is 58.44 g/mol
Therefore the total grams of NaCl needed is
[tex]m_{NaCl}=58.44*0.006275\\ =0.367\ g[/tex]
