Question

Derive the relation between molarity molality and density

Answer 1

Answer:

Molarity is defined as number of moles divided by the volume (in liters). Molality is defined as the number of moles divided by the mass of the solution (in kg). molality = no. of moles of solute/ wt. of solvent. if density of solvent is given we can find out relation between molality

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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.

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