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A sample of NaCIO_{3}  is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl. The mass of AgCl (in g) obtained will be : (Given : Molar mass of AgCI=143.5 g mol_{-1})

  • Option 1)

    0.35

     

  • Option 2)

    0.41

  • Option 3)

    0.48

  • Option 4)

    0.54

 

Answers (1)

As we learned

NaCIO_{3} \vec{\Delta } 2NaCI +3O_{2}

                                            (0.1bg)

\frac{nNaCI}{2}= \frac{no_{2}}{3}

nNaCI = \frac{0.16}{32}\times \frac{2}{3} = \frac{1}{200}\times \frac{2}{3}=\frac{1}{300}

NaCI \rightarrow AgCI

P_{0}AC  on CI

1\times nNaCI = 1\times AgCI

\frac{1}{300} = nAgCI

the weight of  AgCI=\frac{1}{300}\times \left [ 108+35.5 \right ] = \frac{1}{300}\times 143.5

=0.48g

 

 


Option 1)

0.35

 

Option 2)

0.41

Option 3)

0.48

Option 4)

0.54

Posted by

Vakul

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