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Byju's Answer

Standard XII

Chemistry

Percentage Composition

A sample of ...

Question

A sample of $N a C l O_{3}$ 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 $A g C l = 143.5 g m o l^{- 1}$ ]

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Solution

$2 N a C l O_{3} Δ - \to 2 N a C l + 3 O_{2}$

$1$ moles $3$ moles
$2 \times 58.5 g - - - 3 \times 32 g$
$x ? - - - - - 0.16 g$
Now,
$\frac{x}{2 \times 58.5 g} = \frac{0.16 g}{3 \times 32 g}$

$x = \frac{0.16 \times 2 \times 58.5}{3 \times 32} g$

$= 0.195 g$

Now,
$N a C l (a q) + A g^{+} (a q) \to A g C l (s) + N a^{+} (a q)$
$↓$ $↓$
$58.5 g$ ------------------------------ $143.5 g$
$0.195 g$ ----------------------------- $? (g)$

Now,
$\frac{y}{143.5} = \frac{0.195}{58.5}$

$y = \frac{0.195 \times 143.5}{58.5}$

$= 0.478 g$
$∴$ Mass of $A g C l$ attained $= 0.478 g$

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A sample of NaClO3 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=143.5gmol−1]

A sample of $N a C l O_{3}$ 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 $A g C l = 143.5 g m o l^{- 1}$ ]