The relationship between molality m and mole fraction -A of the solvent is-

Mole fraction of solute, X_B = n_Bn_A + n_B Molality= m = n_BW_A (g)× 1000 = n_Bn_AM_A× 1000 nbsp; ⇒n_B/nA nbsp;= X_B/X_A = (1 X_A)/X_A ⇒ m = 1 X_A/X_A×100 ...

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Standard XI

Chemistry

Mole Fraction

The relations...

Question

The relationship between molality (m) and mole fraction $(χ_{A})$ of the solvent is:

m = \frac{1 - X A}{X A} \times \frac{1000}{M_{A}}

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m = \frac{X A}{1 - X A} \times \frac{1000}{X_{A}}

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m = \frac{X A}{1 - X A} \times \frac{1000}{X B}

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m = \frac{1 - X_{A}}{X A} \times \frac{1000}{X B}

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Solution

The correct option is A $m = \frac{1 - X A}{X A} \times \frac{1000}{M_{A}}$
Mole fraction of solute, $X_{B} = \frac{n_{B}}{n_{A} + n_{B}} Molality= m = \frac{n_{B}}{W_{A} (g)} \times 1000 = \frac{n_{B}}{n_{A} M_{A}} \times 1000 \Rightarrow \frac{n_{B}}{n A} = \frac{X_{B}}{X_{A}} = \frac{(1 - X_{A})}{X_{A}} \Rightarrow m = \frac{1 - X_{A}}{X_{A}} \times \frac{1000}{M_{A}}$

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Similar questions

Q. The relationship between molality (m) and mole fraction

(χ_{A})

of the solvent is:

Q. The relationship between molality (m) and mole fraction

(χ_{A})

of the solvent is:

Molality: It is defined as the number of moles of the solute present in 1 kg of the solvent. It is denoted by 'm'
Molality (m) =

\frac{N u m b e r o f m o l e s o f s o l u t e}{N u m b e r o f k i l o g r a m s o f t h e s o l v e n t}

Let

w_{A}

grams of the solute of molecular mass

m_{A}

be present in

w_{B}

grams of the solvent, then
Molality (m) =

\frac{w_{A}}{m_{A} \times w_{B}} \times 1000

Relation between mole fraction and Molality :

X_{A} = \frac{n}{N + n} a n d X_{B} = \frac{N}{N + n}

\frac{X_{A}}{X_{B}} = \frac{n}{N} = \frac{M o l e s o f s o l u t e}{M o l e s o f s o l v e n t} = \frac{W_{A} \times m_{B}}{w_{B} \times m_{A}}

\frac{X_{A} \times 1000}{X_{B} \times m_{B}} = \frac{w_{A} \times 1000}{W_{B} \times m_{A}} = m

, or,

\frac{X_{A} \times 1000}{(1 - X_{A}) m_{B}} = m

If the ratio of the mole fraction of a solute is changed from

\frac{1}{3}

\frac{1}{2}

in the

800

g of solvent then, the ratio of molality will be :

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Mole Fraction

Standard XI Chemistry

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The relationship between molality (m) and mole fraction (χA) of the solvent is:

The correct option is A m=1−XAXA×1000MAMole fraction of solute, XB=nBnA+nBMolality= m=nBWA(g)×1000=nBnAMA×1000⇒nBnA=XBXA=(1−XA)XA⇒m=1−XAXA×1000MA

The relationship between molality (m) and mole fraction $(χ_{A})$ of the solvent is: