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A compound H(2)X with molar mass of 80 g...

A compound `H_(2)X` with molar mass of 80 g is dissolved in a solvent having density of `0.4 g mL^(-1)`. Assuming no change in volume upon dissolution, what is the molality of 3.2 M solution of the compound ?

Text Solution

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Let the olume of 3.2 M solution be 1000 mL
Since there is no change in volume on dissolution
Volume of the solvent = 1000 mL
Mass of the solvent = Volume `xx` density `(1000 mL xx 0.4 "g mL"^(-1)) = 400 g`
No. of moles of the solute present in the solution = 3.2 mol
`:.` Molality of the sodium `(m)=("No. of moles of the solute")/("Mass of the solvent in Kg")`
`=((3.2 "mol"))/(400//1000kg)=8 "mol kg"^(-1)=8m`
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