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What is the molarityk and molality of a 13% solution (by weight) of sulphric acid with a density of `1.02 mL^(-1)`? To what volume should `100 mL` of this acid be diluted in order to preapre a `1.5 N` solution? |
Answer» Correct Answer - Molality=`1.52 m`;`normality=2.70 V=180 L` Molarity =`(% "by weight" xx "Density" xx 10)/("Molecular weight")` =`(13 xx 1.02 xx 10)/98` =`132.6/98` =`1.35 M` `13%` solution by weight means that `13 g` of solute is dissolved in `87 g` of solvent. Thus, molality =(`("Weight of solute")/("Molecular weight of solute"))/("Weight of solvent")xx1000` =`(13/98)/87xx1000` =`(13 xx 1000)/(98 xx 87)` =`1.52 m` Normality = Molarity `xx` Ew factor `:. N=1.35 xx 2=2.70N` For dilution `N_(1)V_(1)=N_(2)V_(2)` `100 xx 2.70N =1.5 N xx V_(2)` `V_(2)=180 mL` So, the acid should be diluted upto `180 mL` to prepare `1.5 N` solution. |
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