Question

# A mixture of NaI and NaCl on reaction with H2SO4 gave Na2SO4 equal to weight of original mixture taken. The percentage of NaI in the mixture is

A
82.38
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B
26.38
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C
62.38
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D
28.38
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Solution

## The correct option is B 26.38Mixture of NaI and NaClLet NaI be x gmNaCl be y gm2 NaI+H2SO4→Na2SO4+2HI2 NaCl+H2SO4→Na2SO4+2HClMNaI=1.50 gm, MNaCl=58.5 gm, MNa2SO4=142 gm2 mole NaI required to give 1 mole Na2SO4 same for NaCl:2 mole give 1 mole Na2SO4.Now mass of Na2SO4=wt− of original mixture =(x+y 1 gm)nNa2SO4=(x150+y58.5)×12∴ Mass of Na2SO4=(x150+y58.5)×12×142=x+y71x150+71y58.5=x+y0.473x+1.21y=x+y⇒ 0.52 x=0.21 y∴ y of NaI=xx+y×100xx+0.5270.21x×100=0.210.737×100=28.4 gm

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