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Standard X

Chemistry

Mass-Mass Relation

10 g of a mix...

Question

10 g of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g. of barium sulphate is precipitated according to the equation :-
$N a_{2} S O_{4} + B a C l_{2} ⟶ B a S O_{4} + 2 N a C l$ .
The percentage of sodium sulphate in the original mixture is:
$[N a = 23, O = 16, S = 32, B a = 132]$

4.26 %

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42.6 %

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52.63 %

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5.26 %

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Solution

The correct option is D 42.6 %
$N a_{2} S O_{4} + B a C l_{2} ⟶ B a S O_{4} + 2 N a C l$
Molecular weight of $B a S O_{4} = 137 + 32 + 64 = 233$
Molecular weight of $N a_{2} S O_{4} = 2 * 23 + 32 + 64 = 142$
$233 g$ of $B a S O_{4}$ is obtained from $142 g$ of $N a_{2} S O_{4}$
$6.99 g$ of $B a S O_{4}$ will be obtained from $\frac{142}{233} \times 6.99$ = $4.26 g$ of $N a_{2} S O_{4}$
Percentage of $N a_{2} S O_{4}$ in $10 g$ of mixture = $\frac{4.26}{10} \times 100$ = $42.6$ %

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Similar questions

10 g of a mixture of barium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is precipitated according to the equation given below:

Na₂SO₄ + BaCl₂ → BaSO₄ +2NaCl

[O = 16, Na = 23, S = 32, Ba = 137]

Calculate the percentage of sodium sulphate in the original mixture.

Q. 1.42 g of sodium sulphate is dissolved in 100 mL water. 4.16 g of barium chloride is dissolved in another 100 mL water. The amount of precipitate of barium sulphate upon mixing of the two solutions is:

N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 N a C l

(Atomic mass: Na = 23, S = 32, O =16, Ba = 137, Cl = 35.5)

N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 N a C l

(Atomic mass: Na = 23, S = 32, O =16, Ba = 137, Cl = 35.5)

Q. What weight of

B a C l_{2}

would react with

24.4 g

of sodium sulphate to produce

46.6 g

of barium sulphate and

23.4 g

of sodium chloride?
(Given,

B a C l_{2} + N a_{2} S O_{4} \to B a S O_{4} + 2 N a C l

)

Q. What weight of

B a C l_{2}

(in grams) would react with

24.4 g

of sodium sulphate to produce

46.6 g

of barium sulphate and

23.4 g

of sodium chloride?
(Given,

B a C l_{2} + N a_{2} S O_{4} \to B a S O_{4} + 2 N a C l

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