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(B) Mass of solution $= 100 \ g$.Mass of $H_2SO_4 = 13 \ g$.Volume of solution $= \frac{100 \ g}{1.09 \ g/mL} = 91.74 \ mL = 0.09174 \ L$.Moles of $H_

Vedclass · Accepted Answer

$1.445 \ M, 2.89 \ N$

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(B) Mass of solution $= 100 \ g$.Mass of $H_2SO_4 = 13 \ g$.Volume of solution $= \frac{100 \ g}{1.09 \ g/mL} = 91.74 \ mL = 0.09174 \ L$.Moles of $H_2SO_4 = \frac{13 \ g}{98 \ g/mol} = 0.1326 \ mol$.Molarity $(M) = \frac{	ext{moles of solute}}{	ext{volume of solution in } L} = \frac{0.1326}{0.09174} \approx 1.445 \ M$.Normality $(N) = 	ext{Molarity} 	imes n	ext{-factor}$.For $H_2SO_4$,$n	ext{-factor} = 2$.Normality $= 1.445 	imes 2 = 2.89 \ N$.

The density of a solution containing $13 \%$ by mass of $H_2SO_4$ is $1.09 \ g/mL$. The molarity and normality of the solution are respectively :-

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