The mole fraction of a solvent in aqueou...

The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality `("in mol kg"^(-1))` of the aqueous solution is:

A

13.88

B

`13.88xx10^(-2)`

C

`13.88xx10^(-3)`

D

`13.88xx10^(-1)`

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The correct Answer is:

A

`m=(x_("solution")xx1000)/(x_("solvent")"x molar mass of solvent")=13.88`

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