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

Apr 25, 2018

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

#### Explanation:

In the given question for $A g C l$ solution ;

THEORETICALLY :-

molality $= 2 m$

$\Rightarrow 2 m =$(Moles of $A g C l$)/(Mass of ${H}_{2} O$ in $k g$)

$\Rightarrow$ Moles of $A g C l$ $= 2 m \times$(Mass of ${H}_{2} O$ in $k g$)

$\Rightarrow$ Moles of $A g C l$ $= 2 m \times$(Mass of ${H}_{2} O$ in $g$)$\times 1000$

We know that the Molecular weight of $A g C l$ is $143.32 g$ :-

$\therefore$ The amount (grams) of $A g C l$ needed

$=$Moles of $A g C l \times 143.32 g$

= {2mxx(Mass of ${H}_{2} O$ in $g$)$\times 1000$}$\times 143.32 g$

NOTE : Actually (EXPERIMENTALLY)this is NOT possible because $A g C l$ is insoluble in Water.

Hope is helps :)