JEE MAIN - Chemistry (2018 - 15th April Morning Slot - No. 14)
A sample of $$NaCl{O_3}$$ 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 $$AgCl=143.5$$ $$g$$ $$mo{l^{ - 1}}$$)
$$0.35$$
$$0.41$$
$$0.48$$
$$0.54$$
Explanation
NaClO3 $$\buildrel \Delta \over
\longrightarrow $$ 2NaCl + 3O2
Here O2 produced = 0.16 g
$$\therefore\,\,\,\,$$ No of moles of O2 = $${{0.16} \over {32}}$$ = 5 $$ \times $$10$$-$$3
Let no of moles of NaCl
$$\therefore\,\,\,\,$$ $${{{}^nNaCl} \over 2}$$ = $${{^n{O_2}} \over 3}$$
$$ \Rightarrow $$ $$\,\,\,\,$$ $$^nNaCl$$ = $${2 \over 3}$$ $$ \times $$ 5 $$ \times $$ 10$$-$$3 = $${1 \over {300}}$$
NaCl + Ag+ $$ \to $$ AgCl + Na+
From here, you can see
no of moles in NaCl = no of moles in AgCl
$$\therefore\,\,\,\,$$ no of moles in AgCl = $${1 \over {300}}$$
$$\therefore\,\,\,\,$$ Mass of AgCl = 143.5 $$ \times $$ $${1 \over {300}}$$ = 0.48 g
Here O2 produced = 0.16 g
$$\therefore\,\,\,\,$$ No of moles of O2 = $${{0.16} \over {32}}$$ = 5 $$ \times $$10$$-$$3
Let no of moles of NaCl
$$\therefore\,\,\,\,$$ $${{{}^nNaCl} \over 2}$$ = $${{^n{O_2}} \over 3}$$
$$ \Rightarrow $$ $$\,\,\,\,$$ $$^nNaCl$$ = $${2 \over 3}$$ $$ \times $$ 5 $$ \times $$ 10$$-$$3 = $${1 \over {300}}$$
NaCl + Ag+ $$ \to $$ AgCl + Na+
From here, you can see
no of moles in NaCl = no of moles in AgCl
$$\therefore\,\,\,\,$$ no of moles in AgCl = $${1 \over {300}}$$
$$\therefore\,\,\,\,$$ Mass of AgCl = 143.5 $$ \times $$ $${1 \over {300}}$$ = 0.48 g
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