@ Moderators@ Stars @ Best UsersDefine Molarity and Molality Derive th

1.	@ Moderators@ Stars @ Best UsersDefine Molarity and Molality Derive the relation between them
Answer» ♣ 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. $\bf m = \dfrac{n_2 }{W_1(in \:kg)} \\ \\ \bf m = \dfrac{n_2 }{ W_1(in \:g)} \times 1000$ ----------------------- ♣ Relation b/W Molarity and Molality :- $\large \purple{ \underline {\boxed{{\bf \frac{d}{ M } = \frac{1}{m} + \frac{M_2}{1000} }}}}$ ----------------------- ♣ 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. $\LARGE\red{\mathfrak{ \text{W}hich \:\:is\:\: the\:\: required} }\\ \Huge \red{\mathfrak{ \text{ A}nswer.}}$ -----------------------

@ Moderators@ Stars @ Best UsersDefine Molarity and Molality Derive the relation between them

Answer»

♣ 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.

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