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Find the molarity strength of a sample which is prepared by mixing $\mathrm{7000 \mathrm{~mL}}$ of $\mathrm{0.5 \mathrm{M} \, \, \mathrm{H}_2 \mathrm{SO}_4}$ (specific gravity $\mathrm{=1.03}$ ) and $\mathrm{4 \mathrm{~L}}$ of $\mathrm{98 \%(\mathrm{w} / \mathrm{V})}$ of $\mathrm{\mathrm{H}_2 \mathrm{SO}_4}$ (specific gravity $\mathrm{=1.84}$ ). Also, the specific gravity of the resultant mixture is 1.4 and assume there is no loss of mass during mixing
Option: 1
0.41 m

Option: 2
0.85 m

Option: 3
4.2 m

Option: 4
0.5 m

Answers (1)

best_answer

since there is no loss of mass due to mixing
$\therefore$ mass of $\mathrm{7 L}$ of $\mathrm{0.5 \mathrm{M\, \, H}_2 \mathrm{SO}_4+}$ mass of $\mathrm{4 \mathrm{~L}}$ of $\mathrm{98 \%, \mathrm{H}_2 \mathrm{SO}_4 =}$ mass of resulting solution

$\mathrm{\therefore \text { mass of solution }=7000 \times 1.03+4000 \times 1.84}$

$=14570$

$\mathrm{\text { Hence, volume of solution }=\frac{14570}{1.4}}$

$\mathrm{=10407.14 \mathrm{~mL}}$

Moles of $\mathrm{\mathrm{H}_2 \mathrm{SO}_4}$ present in 1st kind of

$\mathrm{ \begin{aligned} \text { acid solution } & =7000 \times 0.5 \times 10^{-3} \\ & =3.5 \text { moles } \end{aligned} }$
Amount of. $\mathrm{\mathrm{H}_2 \mathrm{SO}_4}$ present in $\mathrm{4 \mathrm{~L}}$ of

second kind of solution $\mathrm{=\frac{980 \times 4}{98} }$

$\mathrm{=40 \text { moles }}$

$\therefore$ Total moles present in resulting

$\mathrm{ \begin{aligned} \text { mixture } & =3.5+40 \\ & =43.5 \text { moles } \end{aligned} }$

$\therefore$ concentration of resulting solution

$\mathrm{=\frac{43.5 \times 100}{10407.14}}$
$\mathrm{=0.41 \mathrm{M}}$

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Pankaj

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