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`5g` of compound `A` was dissolved in `100g` of water at `303K`. The vapour pressure of the solution was found to be `4.16` kilopascal. If the vapour pressure of pure water is `4.24 kPa` at this temperature, what is the molecular mass of `A`? |
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Answer» Given Mass of solute `(A)=5g` Mass of solvent (water)`=100g` vapour pressure of the solution`=4.16` killopasical vapour pressure of solvent `=4.24kPa` Asked molecular mass of solute `=?` Formulae: `(P_(A)^(0)-P_(A))/(P_(A)^(0))=x_(B)=(W_(B)/(M_(B)))/(W_(A)/M_(A)+W_(B)/(M_(B)))` Explanation : `P_(A)^(0)=` vapour pressure of pure solvent `P_(A)=` vapour pressure of solution `X_(B)=` Mole fraction of solute `W_(B)=` Molecular weight solute `W_(A)=` Mass of solvent `M_(A)=` Molecular weight of solvent Substitution `&` Calculation `(4.21-4.16)/(4.24)=(5)/(M_(B)/(100/(18)+(5)/(M_(B))))rArr(0.08)/(4.23)=(5)/(M_(B)/(100/(18)+(5)/(M_(B))))rArr(0.08xxM_(B))/(M_(B))+0.08xx(100)/(M_(B))xx4.24` `rArr(0.40)/(M_(B))+(8)/(18)=(21.20)/(M_(B))rArr(20.80)/(M_(B))=(8)/(18)` `rArr M_(B)=(20.80xx18)/(8)=46.8g mol^(-1)`. |
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