Question

to find Ce^4+content of a solid ,4.37gwere dissolved and treated with excess iodateto precipitate ce(IO3)4.the precipitated collected,washed well,dried ignited toproduce0.104g of CeO2(172.114)wt was the weight percent of Ce inthe original solid​

Answer 1

Answer:

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Answer 2

Given:

4.37 $gm$ solid dissolved and reacted with iodate to precipitate $Ce$. After collected, it dried and produced 0.104 $gm$ of $CeO_{2}$

To Find:

Percentage of $Ce$ in the solid

Solution:

Since,

Mole is ratio of given weight to molecular weight

Molarity is the no. of moles containing in the volume of the solution

Molality is the no. of moles containing in the weight of solvent

Mole of $Ce$ = Mole of  $CeO_{2}$

$\dfrac{Weight (Ce)}{Atomic weight(Ce)} = \dfrac{Precipate (gm)}{ Formula weight}$

$\dfrac{Weight (Ce)}{140.1} = \dfrac{104}{172.114}$

$Ce (mg) = 84.66$

Percentage of $Ce$ = $\dfrac{84.66}{4370}$

                              = 194 × 100

                              = 1.94%

1.94 % is the answer of this question