AgNO3 10.0 g 1.0×10^(-2) M NaCl 50. mL AgCl

AgNO3 10.0 g 1.0×10^(-2) M NaCl 50. mL AgCl

 

 

Calculate the mass of AgCl formed,

and the concentration of silver ion remaining in solution,

when 10.0 g of solid AgNO3 is added to

50. mL of 1.0×10^(-2) M NaCl solution.

Assume there is no volume change upon addition of the solid.

 

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(1.0×10^(-2) mol/L) (50./1000 L) = 0.00050 mol NaCl

( 참고: MV = mol https://ywpop.tistory.com/7787 )

 

 

 

AgNO3의 몰질량 = 169.87 g/mol 이므로,

10.0 g / (169.87 g/mol) = 0.05887 mol AgNO3

( 참고: n = W/M https://ywpop.tistory.com/7738 )

 

 

 

NaCl + AgNO3 → AgCl + NaNO3

NaCl(aq) + AgNO3(s) → AgCl(s) + NaNO3(aq)

 

NaCl : AgNO3 : AgCl = 1 : 1 : 1 계수비(= 몰수비)

---> 몰수가 작은 NaCl이 한계 반응물.

( 참고: 한계 반응물 https://ywpop.tistory.com/3318 )

 

 

 

AgCl의 몰질량 = 143.32 g/mol 이므로,

0.00050 mol × (143.32 g/mol) = 0.07166 g AgCl

---> 답: 0.072 g AgCl

 

 

 

0.05887 – 0.00050 = 0.05837 mol Ag^+

---> 용액 중 남아있는 Ag^+의 몰수

 

 

 

몰농도 = 용질 mol수 / 용액 L수

= 0.05837 mol / (50./1000 L) = 1.1674 M Ag^+

---> 답: 1.2 M Ag^+