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Equivalent Mass

21.2 g sample...

Question

$21.2 g$ sample of impure $N a_{2} C O_{3}$ is dissolved and reacted with a solution of $C a C l_{2}$ , the weight of $C a C O_{3}$ formed was $10.0 g$ . Which of the following statements is/are correct?

The percentage purity of

N a_{2} C O_{3}

50 %

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The percentage purity of

N a_{2} C O_{3}

60 %

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The number of moles of

N a_{2} C O_{3} = C a C O_{3} = 0.1 mol

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The number of moles of

N a C l

formed is

0.1 mol

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Solution

The correct option is C The number of moles of $N a_{2} C O_{3} = C a C O_{3} = 0.1 mol$
Given: Mass of $N a_{2} C O_{3} = 21.2 g$
Mass of $C a C O_{3} = 10 g$
Molar mass of $N a_{2} C O_{3} = (23 \times 2) + 12 + (16 \times 3) = 106 g$
Molar mass of $C a C O_{3} = 40 + 12 + (16 \times 3) = 100 g$
Let, mass of impurity $= x g$
Balanced chemical reaction is
$N a_{2} C O_{3} + C a C l_{2} \to C a C O_{3} + 2 N a C l$
From stoichiometry:
Moles of $N a_{2} C O_{3} =$ Moles of $C a C O_{3}$
Since, moles = mass/Molar mass
$\frac{21.2 - x}{106} = \frac{10}{100}$
$212 - 10 x = 106$
$x = 10.6 g$
$%$ purity of $N a_{2} C O_{3} = \frac{Mass of pure N a_{2} C O_{3}}{Mass of impure N a_{2} C O_{3}} \times 100$
$= \frac{10.6}{21.2} \times 100 = 50 %$
Moles of $N a_{2} C O_{3} = \frac{21.2 - x}{106} = \frac{21.2 - 10.6}{106} = 0.1$
Moles of $C a C O_{3} = \frac{10}{100} = 0.1$
From stoichiometry,
$\frac{moles of N a_{2} C O_{3}}{1} = \frac{moles of N a C l}{2}$
moles of $N a C l = 0.1 \times 2 = 0.2$
So, options a and c are correct.

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Similar questions

21.2 g

sample of impure

N a_{2} C O_{3}

is dissolved and reacted with a solution of

C a C l_{2}

, the weight of precipitate of

C a C O_{3}

10.0 g

. Which of the following statements is/are correct ?

21.2 g

sample of impure

N a_{2} C O_{3}

is dissolved and reacted with a solution of

C a C l_{2}

, the weight of

C a C O_{3}

formed was

10.0 g

. Which of the following statements is/are correct?

21.2 g

sample of impure

N a_{2} C O_{3}

is dissolved and reacted with a solution of

C a C l_{2}

, the weight of precipitate of

C a C O_{3}

10.0 g

. Which of the following statements is/are correct ?

6.2

g of a sample containing

N a_{2} C O_{3}, N a H C O_{3}

and non-volatile inert impurity on gentle heating loses

5 %

of its weight due to reaction:

2 N a H C O_{3} \to N a_{2} C O_{3} + H_{2} O + C O_{2}

. Residue is dissolved in water and formed

100 m L

solution and its

10 m L

portion requires

7.5 m L

0.2 M

aqueous solution of

B a C l_{2}

for complete precipitation of carbonates. Determine weight (in g) of

N a_{2} C O_{3}

in the original sample.

Q. Which of the following statements is/are correct?

1.0

g mixture of

C a C O_{3} (s)

and glass beads liberate

0.22

g of

C O_{2}

upon treatment with excess of

H C l

. Glass does not react with

H C l

C a C O_{3} + 2 H C l \to C O_{2} + H_{2} O + C a C l_{2}

[Molecular weight of

C a C O_{3} = 100

g/mol,

C O_{2} = 44

g/mol and atomic weight of

C a = 40

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21.2 g sample of impure Na2CO3 is dissolved and reacted with a solution of CaCl2, the weight of CaCO3 formed was 10.0 g. Which of the following statements is/are correct?

$21.2 g$ sample of impure $N a_{2} C O_{3}$ is dissolved and reacted with a solution of $C a C l_{2}$ , the weight of $C a C O_{3}$ formed was $10.0 g$ . Which of the following statements is/are correct?