How many grams of AgCl are needed to make a 2m solution?

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Answer 1 · 2018-04-25T22:50:19

Molality = (Number of moles of solute)/(Mass of solvent in $kg$ )

In the given question for $AgCl$ solution ;

THEORETICALLY :-

molality $= 2m$

$rArr 2m=$ (Moles of $AgCl$ )/(Mass of $H_2O$ in $kg$ )

$rArr$ Moles of $AgCl$ $= 2mxx$ (Mass of $H_2O$ in $kg$ )

$rArr$ Moles of $AgCl$ $= 2mxx$ (Mass of $H_2O$ in $g$ ) $xx1000$

We know that the Molecular weight of $AgCl$ is $143.32 g$ :-

$:.$ The amount (grams) of $AgCl$ needed

$=$ Moles of $AgClxx143.32 g$

$= {2mxx$ (Mass of $H_2O$ in $g$ ) $xx1000$ } $xx143.32 g$

NOTE : Actually (EXPERIMENTALLY)this is NOT possible because $AgCl$ is insoluble in Water.

Hope is helps :)