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Standard IX

Chemistry

Introduction to Acid, Base and Salts

0.56 g of lim...

Question

$0.56 g$ of limestone was treated with oxalic acid to give $C a C_{2} O_{4}$ . The precipitate decolorised $45 m L$ of $0.2 N K M n O_{4}$ in acid solution. Calculate the % of $C a O$ in limestone.

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Solution

$45 m L$ of $0.2 N K M n O_{4} = 45 m L$ of $0.2 N C a C_{2} O_{4}$
$= 45 m L$ of $0.2 N C a C O_{3}$
$= 45 m L$ of $0.2 N C a O$
Mass of $C a O = \frac{N \times E \times V}{1000} = \frac{0.2 \times 28 \times 45}{1000} = 0.252 g$
% of $C a O$ in limestone $= \frac{0.252}{0.56} \times 100 = 45$ .

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Similar questions

0.56

g of limestone was treated with oxalic acid to give

C a C_{2} O_{4}

. The precipitate decolourised

50 m L

\frac{M}{25}

K M n O_{4}

in acidic medium. If

X

percentage of

C a O

is present in limestone, then the value of

X

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 then the calcium is precipitated as

C a C_{2} O_{4}

. After filtering and washing the precipitate, it requires

40 m L

0.25 N K M n O_{4}

solution acidified with

H_{2} S O_{4}

to titrated it as,

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

The percentage of

C a O

in the sample is:

Q. 0.56g of lime (CaO) was treated with 100ml of 0.5M oxalic acid is give ppt of

C a C_{2} O_{4}

. The filtrate required V ml of 0.2M NaOH for complete neutralization. The value of V is: (atomic Weight of Ca=40, O=16)

Q. A

1.5

g sample containing oxalic acid and some inert impurity was dissolved in enough water and volume was made up to

250

mL. A

20

mL portion of this solution was then mixed with

30

mL of an alkali solution. The resulting solution was then treated with stoichiometric amount of

C a C l_{2}

just needed for precipitation of oxalate as

C a C_{2} O_{4}

solution was filtered off and filtrate was finally titrated against

0.1 M H C l

solution.

8.0

mL of acid was required to reach the equivalence point. At last the above neutral solution was treated with excess of

A g N O_{3}

solution and

A g C l

obtained was washed, dried and weighed to be

0.4305

g. Determine the mass percentage of oxalic acid in the original sample.

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0.56 g of limestone was treated with oxalic acid to give CaC2O4. The precipitate decolorised 45 mL of 0.2 N KMnO4 in acid solution. Calculate the % of CaO in limestone.

45 mL of 0.2 N KMnO4=45 mL of 0.2 N CaC2O4=45 mL of 0.2 N CaCO3=45 mL of 0.2 N CaOMass of CaO=N×E×V1000=0.2×28×451000=0.252 g% of CaO in limestone =0.2520.56×100=45.

$0.56 g$ of limestone was treated with oxalic acid to give $C a C_{2} O_{4}$ . The precipitate decolorised $45 m L$ of $0.2 N K M n O_{4}$ in acid solution. Calculate the % of $C a O$ in limestone.

$45 m L$ of $0.2 N K M n O_{4} = 45 m L$ of $0.2 N C a C_{2} O_{4}$
$= 45 m L$ of $0.2 N C a C O_{3}$
$= 45 m L$ of $0.2 N C a O$
Mass of $C a O = \frac{N \times E \times V}{1000} = \frac{0.2 \times 28 \times 45}{1000} = 0.252 g$
% of $C a O$ in limestone $= \frac{0.252}{0.56} \times 100 = 45$ .