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What mass of $95 \%$ pure $\mathrm{CaCO}_{3}$ will be required to neutralise

$50 \mathrm{~mL}$ of $0.5 \mathrm{MHCl}$ solution according to the following reaction?

$\mathrm{CaCO}_{3(s)}+2 \mathrm{HCl}_\mathrm{{(a q)}} \rightarrow \mathrm{CaCl}_{2(\text { aq })}+\mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(1)}$
[Calculate upto second place of decimal point]
Option: 1
$1.32 \mathrm{~g}$

Option: 2
$3.65 \mathrm{~g}$

Option: 3
$9.50 \mathrm{~g}$

Option: 4
$1.25 \mathrm{~g}$

Answers (1)

best_answer

Give reaction

$\mathrm{CaCO}_{3(s)}+2 \mathrm{HCl}_{\text {(aq) }} \rightarrow \mathrm{CaCl}_{2 \text { (aq) }}+\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

From above,

For 2 mole of $\mathrm{HCl}$ = 1 mole of required

$\mathrm{1\; mole\; of\; \mathrm{HCl}=\frac{1}{2}\; mole\; of \;\mathrm{CaCO}_{3}}\\$

$(50 \mathrm{ml} \times 0.5 \mathrm{M}) \mathrm{HCl}=\frac{50 \times 0.5}{2} \mathrm{m\;mole}\; \mathrm{of}\; \mathrm{CaCO}_{3}\\$

$\mathrm{=12.5 m\;mole\; of \;\mathrm{CaCO}_{3}}\\$

$=12.5 \times 10^{-3} \times 100$

$\mathrm{=1.25 \mathrm{~g}\; of\; \mathrm{CaCO}_{3}}$ required

For some of $\mathrm{0.5M\;HCl}$ solution $\mathrm{1.25g}$ of $\mathrm{CaCO_{3}}$ required

But $\mathrm{CaCO_{3}}$ is $\mathrm{85%}$ so,

$\mathrm{FaCO}_{3}\mathrm{{\;is \;95 \%\;so}}$ ,

$\mathrm{\% \text { Purity } =\frac{\text { mass of pure }CaCO_{3}}{\text { Total Mass of impure } CaCO_{3}} \times 100} \\$

$\mathrm{95 =\frac{1.25}{x} \mathrm{~g}} \\$

$\mathrm{x =\frac{1.25}{95} \mathrm{~g}=1.32 \mathrm{~g}}$

Hence the correct option is 1.

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Gunjita

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