Question

derive the relation between molarity and molality..

Answer 1

For this derivation see attachment....

Answer 2

♣ Definitions :-

♠》 Molarity (M) :-

It is the number of moles of solute dissolved in one litre of solution.

$\bf M = \dfrac{n_2 }{V_{sol.}(in \:L)} \\ \\ \bf M= \dfrac{n_2 }{V_{sol.}(in \:mL)} \times 1000$

♠》 Molality (m) :-

It is the number of moles of solute dissolve in one kg of Solvent.

$\bf m = \dfrac{n_2 }{W_1(in \:kg)} \\ \\ \bf m = \dfrac{n_2 }{ W_1(in \:g)} \times 1000$

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♣ Relation b/w Molarity and Molality :-

$\large \purple{ \underline {\boxed{{\bf  \frac{d}{ M }  =  \frac{1}{m} + \frac{M_2}{1000} }}}}$

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♣ Derivation :-

We have ,

$\sf m = \dfrac{n_2 }{ W_1(in \:g)} \times 1000 \: \: \:. \: . \: . \: \{i \} \\ \\ \bf and \\ \\ \sf M = \dfrac{n_2 }{V_{sol.}(in \:mL)} \times 1000 \: \: \: .\: . \: . \: \{ii \}$

Dividing {i} by {ii} We Get,

$\sf\dfrac{m}{ M} = \dfrac{V_{sol.}}{W_1} \\ \\ \sf = \frac{W_{sol.}/d}{W_1} \\ \\ \sf = \frac{W_1 + W_2}{d.W_1}$

$\sf = \dfrac{1}{d} \bigg \{ \dfrac{W_2}{W_1} + 1 \bigg \} \\ \\ \sf = \dfrac{1}{d} \bigg \{ \frac{n_2}{n_2} \times \dfrac{W_2}{W_1} + 1\bigg \} \\ \\ \sf = \frac{1}{d} \bigg \{ \dfrac{n_2}{W_2 /M_2} \times \dfrac{W_2}{W_1} + 1 \bigg \} \\ \\ \sf = \frac{1}{d} \bigg \{ \frac{n_2.M_2}{W_1} + 1\bigg \}\\\\ \sf = \dfrac{1}{d} \bigg \{ \dfrac{m. M_2 }{1000} + 1 \bigg \}$

$\large :\longmapsto \purple{ \underline {\boxed{{\bf  \frac{d}{ M }  =  \frac{1}{m} + \frac{M_2}{1000} }}}}$

♣ Notations Used :-

W = Mass

M = Molar Mass

1 = Solvent

2 = Solute

n = number of moles

d = density in g/mL.

$\LARGE\red{\mathfrak{  \text{W}hich \:\:is\:\: the\:\: required} }\\ \Huge \red{\mathfrak{ \text{ A}nswer.}}$

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