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1.0 xx 10^(-3) mol of Ag^(+) and 1.0 xx ...

`1.0 xx 10^(-3)` mol of `Ag^(+)` and `1.0 xx 10^(-3)` mol of `CrO_(4)^(2-)` react together to form solid `Ag_(2)CrO_(4)`. What is the amount of `Ag_(2)CrO_(4) = 332)`

A

0.268gm

B

0.166gm

C

0.212gm

D

1.66gm

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The correct Answer is:
B
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1.0L of solution which was in equilibrium with solid mixture of AgCI and Ag_(2)CrO_(4) was found to contain 1 xx 10^(-4) moles of Ag^(+) ions, 1.0 xx 10^(-6) moles of CI^(-) ions and 8.0 xx 10^(-4) moles of CrO_(4)^(2-) ions. At constant volume, Ag^(+) ions are added slowly to the above mixture till 8.0 xx 10^(-7) moles of AgCI got precipitated. How many moles of Ag_(2)CrO_(4) were also precipitated ? Given your answer after multiplying with 10^(6) .

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When 15mL of 0.05M AgNO_(3) is mixed with 45.0mL of 0.03M K_(2)CrO_(4) , predict whether precipitation of Ag_(2)CrO_(4) occurs or not? K_(sp) of Ag_(2)CrO_(4) = 1.9 xx 10^(-12)

Which is the correct representation for the solubility product constant of Ag_(2)CrO_(4) ?

The [Ag^(+)] ion in a saturated solution of Ag_(2)CrO_(4) at 25^(@)C is 1.5xx10^(-4)M . Determine K_(SP) of Ag_(2)CrO_(4) at 25^(@)C .

At 25^oC , if the concentration of Ag^+ ion is 1.5xx10^-4 mol/L in the saturated solution of Ag_2CrO_4 , then solubility product of Ag_2CrO_4 is

The volume of 0.1M AgNO_(3) which is required by 10 ml of 0.09 M K_(2)CrO_(4) to precipitate all the chromate as Ag_(2)CrO_(4) is

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