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Question

64g of a mixture of NaCl and KCl were treated with concentrated sulphuric acid. The total mass of metal sulphates obtained was found to be %76g. What are the mass percents of NaCl in the mixture. The reactions are 2NaCl+H2SO4→Na2SO4+2HCl; 2KCl+H2SO4→K2SO4+2HCl;

A
42.89%
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B
84.9%
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C
31.5%
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D
63.1%
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Solution

The correct option is B 84.9%
Given,
Weight of NaCland KClmixture = 64 gm
Weight of metal sulphates, Na2SO4and K2SO4 = 76 gm

The reactions will be:
2NaCl+H2SO4Na2SO4+2HCl
2KCl+H2SO4K2SO4+2HCl

Now, let mass of NaCl be x gm.
Therefore, moles of NaCl will be =x58.5
And moles of KClwill be =64x74.5
From the above reactions we can write that

Moles of NaCl*1 = Moles of Na2SO4*2, or
Moles of Na2SO4 = Moles of NaCl* ½ ...(i)

Similarly, for KCl and K2SO4 we can write
Moles of K2SO4*1 = Moles of KCl* ½ ...(ii)

It was given that the total weight of the metal sulphates was 76 gm, therefore from equation (i)and (ii)we get

Weight of Na2SO4 + Weight of K2SO4 = 76 gm
12×x58.5×142+12×64x74.5×174=76

1.21×74.41.16x=76

Solving the above equation for xwe get,
x=27.45

Now, percentage mass of NaCl can be given as:
=givenmassmolarmass×100
=27.4564×100=42.89%

Hence, the mass percent of NaCl in the mixture is 42.89%

Therefore, the correct option is A.

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