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300 \mathrm{~mL} of solution was made by taking 1.15 \mathrm{~g} of a sample of \mathrm{NaCl} having \mathrm{\mathrm{NaNO}_3} as an impurity. Out of the prepared volume, 25 \mathrm{~mL} of the solution needed 17.95 \mathrm{~mL} of \mathrm{M} / \mathrm{10} \quad \mathrm{AgNO}_3 solutions. What will be the composition in \mathrm{g} / \mathrm{L}.

Option: 1

\mathrm{0.33 \mathrm{~g} / \mathrm{L}}


Option: 2

\mathrm{3.33 \mathrm{~g} / L}


Option: 3

\mathrm{0.66 \mathrm{~g} / \mathrm{L}}


Option: 4

\mathrm{6.66 \, \mathrm{g/L}


Answers (1)

best_answer

Let in 1.15 \mathrm{~g} sample, \mathrm{NaCl} present \mathrm{=x \mathrm{~g}}

\mathrm{ \therefore} in the given sample,\mathrm{ \mathrm{NaNO}_3} present \mathrm{ =(1.15-x) \mathrm{g}}

\mathrm{ 300 \mathrm{~mL}} of resulting solution contains \mathrm{ x \mathrm{~g}\, \, NaCl. }

\mathrm{ \therefore 25 \mathrm{~mL}} of solution contain \mathrm{ =\frac{x}{10} \times \frac{1}{58.5}}

\mathrm{ 17.95 \mathrm{~mL}} of \mathrm{ \mathrm{M} / 10 \quad \mathrm{AgNO}_3=17.95 \times \frac{1}{10} \times 10}

Moles of \mathrm{\mathrm{AgNO}_3=1.795 \times 10^{-3} \, \, moles}

\mathrm{\mathrm{AgNO}_3+\mathrm{NaCl} \rightarrow \mathrm{AgCl}+\mathrm{NaNO}_3}

I mole of \mathrm{\mathrm{AgNO}_3} reacts with 1 formula weight of \mathrm{\mathrm{NaCl}.}

\mathrm{\therefore \, \, 1.795 \times 10^{-3}} moles of \mathrm{\mathrm{AgNO}_3} reacts with \mathrm{1.795 \times 10^{-3}} formula wight of \mathrm{\mathrm{NaCl}.}

 \mathrm{ \begin{aligned} & \therefore \frac{x}{10} \times \frac{1}{58.5}=1.795 \times 10^{-3} \\ & \begin{aligned} x=1.05 \mathrm{~g} \end{aligned} \\ & \begin{aligned} \text { In } 1.15 \mathrm{~g} \text { sample, } \mathrm{NaNO}_3 \text { present } & =1.15-1.05 \\ & =0.1 \mathrm{~g} \\ \text { conc. of } \mathrm{NaCl} \text { in resultant } & =\frac{1.05}{300} \times 1000 \\ & =3.5 \mathrm{~g} / \mathrm{L} \end{aligned} \\ & \text { conc. of } \mathrm{NaNO}_3=\frac{0.1}{300} \times 1000 \end{aligned} }
\mathrm{ =0.33 \mathrm{~g} / \mathrm{L} \text {. } }

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Riya

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