If the ratio of the mole fraction of a solute is changed from 1-3 to 1-2 in the 800 g of solvent then the ratio of molality will be -A- 1- 3B- 3- 1C- 4- 3D- 1- 2

The correct option is D 1 : 2Molality of solution is given by χ_A× 1000(1 χ_A)m_B=m. Consider the molality for two different mole fraction i.e 12,13 of A be m_ ...

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Standard XII

Chemistry

Molality

If the ratio ...

Question

If the ratio of the mole fraction of a solute is changed from $\frac{1}{3}$ to $\frac{1}{2}$ in the 800 g of solvent then the ratio of molality will be :

1 : 3

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3 : 1

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4 : 3

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1 : 2

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Solution

The correct option is D 1 : 2
Molality of solution is given by $\frac{χ_{A} \times 1000}{(1 - χ_{A}) m_{B}} = m$ .
Consider the molality for two different mole fraction i.e $\frac{1}{2}$ , $\frac{1}{3}$ of $A$ be $m_{1}$ and $m_{2}$ respectively.
$m_{1} = \frac{\frac{1}{3} \times 1000}{(1 - (\frac{1}{3})) \times m_{B}}$ ....(i)
$m_{2} = \frac{\frac{1}{2} \times 1000}{(1 - (\frac{1}{2})) \times m_{B}}$ ....(ii)
On dividing equation (i) and (ii) we get
$\frac{m_{1}}{m_{2}} = 1 : 2$

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Similar questions

Q. 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 :

Q. If the mole fraction of a solute is changed from

\frac{1}{4}

\frac{1}{2}

by adding some solute in the

800 g

H_{2} O

solvent, then the ratio of molality of two solutions will be:

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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Molality

Standard XII Chemistry

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If the ratio of the mole fraction of a solute is changed from 13 to 12 in the 800 g of solvent then the ratio of molality will be :

The correct option is D 1 : 2Molality of solution is given by χA×1000(1−χA)mB=m. Consider the molality for two different mole fraction i.e 12,13 of A be m1 and m2 respectively. m1=13×1000(1−(13))×mB....(i) m2=12×1000(1−(12))×mB....(ii) On dividing equation (i) and (ii) we get m1m2=1:2

If the ratio of the mole fraction of a solute is changed from $\frac{1}{3}$ to $\frac{1}{2}$ in the 800 g of solvent then the ratio of molality will be :