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Concentration Terms - An Introduction (w/w etc)

10 mL of a so...

Question

$10$ mL of a solution of $K_{2} S O_{4}$ , on reaction with $B a C l_{2}$ , gave $0.233$ g of white ppt. The concentration of $K_{2} S O_{4}$ solution is:
(Given that the molecular weight of $B a = 137, S = 32, O = 16, K = 39$ )

17.4 g L^{- 1}

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34.8 g L^{- 1}

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8.7 g L^{- 1}

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4.4 g L^{- 1}

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Solution

The correct option is A $17.4 g L^{- 1}$
The balanced chemical equation is $K_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 K C l$
0.233 g of $B a S O_{4}$ corresponds to $\frac{0.233}{233} = 0.001$ moles. They also corresponds to 0.001 moles of $K_{2} S O_{4}$ .
0.001 moles in 10 ml will be equal to 0.1 moles in 1 L.
It is equual to $174 g / m o l \times 0.1 M = 17.4 g / L$ of $K_{2} S O_{4}$ .

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Similar questions

Q. The amount of

B a S O_{4}

formed upon mixing

100

ml of

20.8 %

B a C l_{2}

solution with

50

ml of

9.8 %

H_{2} S O_{4}

solution will be:

(Given that molecular weight of

B a = 137, C l = 35.5, S = 32, H = 1

and

O = 16

g/mol)

Q. 100 mL of 20.8%

B a C l_{2}

solution and 50 mL of 9.8%

H_{2} S O_{4}

solution will form

B a S O_{4}

(B a = 137, C l = 35.5, S = 32, H = 1, O = 16)

B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l

Q. The amount of

B a S O_{4}

formed upon mixing

100

mL of

20.8 %

B a C l_{2}

solution with

50

mL of

9.8 %

H_{2} S O_{4}

solution will be:
[Ba

= 137

, Cl

= 35.5

, S

= 32

, H

= 1

and O

= 16

500

mL of

0.25 M N a_{2} S O_{4}

solution is added to an aqueous solution of

15.0

g of

B a C l_{2}

solution resulting in the formation of white precipitate of

B a S O_{4}

. The moles of

B a S O_{4}

formed is

y \times 10^{- 3}

moles. Then,

y

is (in nearest integer) : [Given that molecular weight of

B a = 137, S = 32, O = 16, N a = 23

g/mol]

0.32

g of an aorganic compound gave

0.233

g of

B a S O_{4}

. Determine the percentage of sulphur in the compound.
(At. mass of

B a = 137, S = 32, O = 16

)

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10 mL of a solution of K2SO4, on reaction with BaCl2, gave 0.233 g of white ppt. The concentration of K2SO4 solution is:(Given that the molecular weight of Ba=137,S=32,O=16,K=39)

The correct option is A 17.4gL−1The balanced chemical equation is K2SO4+BaCl2→BaSO4+2KCl 0.233 g of BaSO4 corresponds to 0.233233=0.001 moles. They also corresponds to 0.001 moles of K2SO4.0.001 moles in 10 ml will be equal to 0.1 moles in 1 L.It is equual to 174g/mol×0.1M=17.4g/L of K2SO4.

$10$ mL of a solution of $K_{2} S O_{4}$ , on reaction with $B a C l_{2}$ , gave $0.233$ g of white ppt. The concentration of $K_{2} S O_{4}$ solution is:
(Given that the molecular weight of $B a = 137, S = 32, O = 16, K = 39$ )