Chemistry, asked by Vikesh30491, 1 year ago

The relation between molarity and molality is given by

Answers

Answered by abhirajpathak9
0
1/m = d/M - Mb/1000

where
m is molality
d is density of solution
M is Molarity
Mb is the molar mass of solute
Answered by XxDangerousQueenxX
1

♣ 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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