1.

The mole fraction of `CH_(3)OH` in an aqueous solution is 0.02 and its density is `0.994 g cm^(-3)`. Determine its molarity and molality.

Answer» Let `x` mole of `CH_(3)OH` and y moles of water be present in solution
Mole fraction of `CH_(3)OH=(x)/(x+y)=0.02`
So, `(y)/(x)=49or(1)/(y)=(1)/(49)`
Molality `= (x)/(18xxy)xx1000=(1000)/(18xx49)=1.13m`
Volume of solution `= ("Total mass")/("density")=(32x+18y)/(0.994)mL`
`=(32x+18y)/(0.994xx1000)"litre"=(32x+18y)/(994)` litre
Molarity `= (x)/(32x+18y)xx994`
`=(994)/(32+18xxy//x)=(994)/(32+18xx49)=1.0875M`.


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