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1.26 g AgNO3 in a 250 mL volumetric flask
A solution is prepared by dissolving 1.26 g AgNO3
in a 250 mL volumetric flask and diluting to the volume.
Calculate the molarity of the silver nitrate solution.
How many millimoles AgNO3 were dissolved?
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AgNO3의 몰질량 = 169.87 g/mol
1.26 g / (169.87 g/mol) = 0.007417 mol AgNO3
( 참고: n=W/M https://ywpop.tistory.com/7738 )
---> 7.42 mmol AgNO3
몰농도 = 용질 mol수 / 용액 L수
( 참고 https://ywpop.tistory.com/3222 )
= 0.007417 mol / 0.250 L
= 0.0297 mol/L
답: 0.0297 M, 7.42 mmol
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