The molarity of 1 L orthophosphoric acid (H_ | vedclass

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(C) Specific gravity (density) $= 1.54 \ g \ cm^{-3}$.Volume of solution $= 1 \ L = 1000 \ mL$.Mass of solution $= 	ext{Density} 	imes 	ext{Volume}

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$11$

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(C) Specific gravity (density) $= 1.54 \ g \ cm^{-3}$.Volume of solution $= 1 \ L = 1000 \ mL$.Mass of solution $= 	ext{Density} 	imes 	ext{Volume} = 1.54 \ g \ cm^{-3} 	imes 1000 \ cm^3 = 1540 \ g$.Since the solution is $70 \%$ pure by weight,the mass of $H_3PO_4 = 0.70 	imes 1540 \ g = 1078 \ g$.Moles of $H_3PO_4 = \frac{	ext{Mass}}{	ext{Molar mass}} = \frac{1078 \ g}{98 \ g \ mol^{-1}} = 11 \ mol$.Molarity $= \frac{	ext{Moles of solute}}{	ext{Volume of solution in } L} = \frac{11 \ mol}{1 \ L} = 11 \ M$.

The molarity of $1 \ L$ orthophosphoric acid $(H_3PO_4)$ having $70 \%$ purity by weight (specific gravity $1.54 \ g \ cm^{-3}$) is $ . . . . . . $ $M$.
(Molar mass of $H_3PO_4 = 98 \ g \ mol^{-1}$)

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The molarity of $1 \ L$ orthophosphoric acid $(H_3PO_4)$ having $70 \%$ purity by weight (specific gravity $1.54 \ g \ cm^{-3}$) is $ . . . . . . $ $M$.(Molar mass of $H_3PO_4 = 98 \ g \ mol^{-1}$)