Question #d1c95

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Question #d1c95

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Answer 1 · 2017-12-08T10:00:27

See the explanation.Not to be confused with molarity.

Explanation:

The mola $l$ $color(red)"l"$ ity of a solution is defined as
$b = \frac{n_{s o l u t e}}{m_{s o l v e n t}}$ $b=(n_(solute))/(m_(solvent))$ , where
$n_{s o l u t e}$ $n_(solute)$ : amount of solute(in moles)
$m_{s o l v e n t}$ $m_(solvent)$ : mass of solvent(in kg)

Since $n_{s o l u t e}$ $n_(solute)$ and $m_{s o l v e n t}$ $m_(solvent)$ are never affected by the temparature, molality is also independent of temparature.

On the other hand, mola $r$ $color(blue)"r"$ ity
$c = \frac{n_{s o l u t e}}{v_{s o l u t i o n}}$ $c=n_(solute)/v_(solution)$ is dependent on the temparature, as $v_{s o l u t i o n}$ $v_(solution)$ , volume of the solution(in L), will increase if the temparature rises and vice versa.

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Question #d1c95

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