1.584 g Mn metal HNO3 1.000 L Mn^2+ A 50.00 mL 1000.0 mL

1.584 g Mn metal HNO3 1.000 L Mn^2+ A 50.00 mL 1000.0 mL

 

 

1.584 g의 순수한 망가니즈 금속을 질산에 녹인 다음

최종 부피가 1.000 L가 되게 하여

Mn^2+ 이온이 들어 있는 저장 용액을 만들었다.

 

 

A stock solution containing Mn^2+ ions was prepared

by dissolving 1.584 g pure manganese metal in nitric acid

and diluting to a final volume of 1.000 L.

The following solutions were then prepared by dilution:

For solution A, 50.00 mL of stock solution was diluted to 1000.0 mL.

For solution B, 10.00 mL of solution A was diluted to 250.0 mL.

For solution C, 10.00 mL of solution B was diluted to 500.0 mL.

Calculate the concentrations of the stock solution and solutions A, B, and C.

 

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Mn의 몰질량 = 54.94 g/mol

1.584 g / (54.94 g/mol) = 0.02883 mol Mn

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

stock solution의 몰농도

= 용질 mol수 / 용액 L수

( 참고 https://ywpop.tistory.com/3222 )

 

= 0.02883 mol / 1.000 L

= 0.02883 mol/L

= 0.02883 M

 

 

 

 

For solution A,

MV = M’V’

( 참고 https://ywpop.tistory.com/2859 )

 

(0.02883 M) (50.00 mL) = (? M) (1000.0 mL)

 

? = (0.02883) (50.00) / (1000.0) = 0.001442 M = 1.442×10^(-3) M

 

 

 

 

For solution B,

(0.001442 M) (10.00 mL) = (? M) (250.0 mL)

 

? = (0.001442) (10.00) / (250.0) = 5.768×10^(-5) M

 

 

 

 

For solution C,

? = (5.768×10^(-5)) (10.00) / (500.0) = 1.154×10^(-6) M