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What is the molarity and molality of a 1...

What is the molarity and molality of a 13% solution (by weight) of sulphric acid with a density of `1.02gmL^(-1)`? To what volume should `100 mL` of this acid be diluted in order to preapre a `1.5 N` solution?

Text Solution

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The correct Answer is:
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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