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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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