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

Answer»

Solution :3.2 molar solution contains 3.2 moles of `H_(2)X` in 1000 mL of the solution. As there is no change in volume on dissolution, volume of solvent= 1000 mL.
`"Mass of solvent= 1000 mL "xx0.4"g mL"^(-1)=400g`
= 0.4 kg
`"MOLALITY"=("3.2 moles")/("0.4 kg")="8 mol kg"^(-1)=8M`


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