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Standard XII

Chemistry

Percentage Composition

The calcium i...

Question

The calcium in a $0.8432$ g sample is precipitated as $C a C_{2} O_{4}$ . The precipitate is washed, ignited to $C a C O_{3}$ and found to weigh $0.3462$ g. Percentage of $C a O$ in the sample is (only integer part)

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Solution

$C a O ⟶ C a C_{2} O_{4} ⟶ C a C O_{3}$
$P u r e C a O i n t h e m i n e r a l = \frac{56}{100} \times 0.3462$ $= 0.1938 g m$
$∴ % o f p u r e C a O = \frac{0.1938}{0.8432} \times 100$ $= 23 %$

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Similar questions

Q. The calcium in a

0.8432

g sample is precipitated as

C a C_{2} O_{4}

. The precipitate is washed, ignited to

C a C O_{3}

and found to weigh

0.3462

g. Percentage of

C a O

in the sample is :

0.518

g sample of limestone is dissolved and calcium is precipitated as calcium oxalate

(C a C_{2} O_{4})

. After filtration and washing the precipitate, it requires

40

mL of

0.05 M K M n O_{4}

solution acidified with sulphuric acid to titrate it. What is the percentage of

C a O

in the limestone? (as nearest integer)

Q. A

0.518

g sample of limestone is dissolved in

H C l

and the the calcium is precipitated as

C a C_{2} O_{4}

. After filtering and washing the precipitate, it requires

40.0

mL of

0.250 N K M n O_{4}

solution acidified with

H_{2} S O_{4}

to titrate it as. The percentage of

C a O

in the sample is :

M n O_{4}^{-} + H^{+} + C_{2} O_{4}^{2 -} ⟶ M n^{2 +} + C O_{2} + 2 H_{2} O

Q. A sample of mixture of

C a C l_{2}

and

N a C l

weighing

5.55

g was treated to precipitate all the

C a

C a C O_{3}

which was then heated and quantitatively converted to

1.68

g of

C a O

. The mass percentage of

C a C l_{2}

in the mixture is;

10

g of a sample of a mixture of

C a C l_{2}

and

N a C l

is treated to precipitate all the calcium as

C a C O_{3}

. This

C a C O_{3}

is heated to convert all the

C a

C a O

and the final mass of

C a O

1.62

g. The percent by mass of

C a C l_{2}

in the original mixture is :

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The calcium in a 0.8432 g sample is precipitated as CaC2O4. The precipitate is washed, ignited to CaCO3 and found to weigh 0.3462 g. Percentage of CaO in the sample is (only integer part)

CaO⟶CaC2O4⟶CaCO3PureCaOinthemineral=56100×0.3462=0.1938 gm∴%ofpureCaO=0.19380.8432×100=23%

The calcium in a $0.8432$ g sample is precipitated as $C a C_{2} O_{4}$ . The precipitate is washed, ignited to $C a C O_{3}$ and found to weigh $0.3462$ g. Percentage of $C a O$ in the sample is (only integer part)

$C a O ⟶ C a C_{2} O_{4} ⟶ C a C O_{3}$
$P u r e C a O i n t h e m i n e r a l = \frac{56}{100} \times 0.3462$ $= 0.1938 g m$
$∴ % o f p u r e C a O = \frac{0.1938}{0.8432} \times 100$ $= 23 %$