InterviewSolution
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InterviewSolution
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The amount of solute (molar mass 60 g `mol^(-1)`) that must be added to 18g of water so that the vapour pressure of water is lowered by 10% isA. 30 gB. 60 gC. 120 gD. 12 g |
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Answer» Correct Answer - B Relative lowering of vapour pressure is given by the formula : `(p^(@)-p_(s))/(p^(@))=(omega_(A))/(m_(A))xx(m_(B))/(omega_(B))` Given, relative lowering of vapour pressure `=(p^(@)-p_(s))/(p^(@))=(10)/(100)` `m_(A)=60, m_(B)=18, omega_(B)=180, omega_(A)=x` `therefore (10)/(100)=(x//60)/(180//18)rArr(1)/(10)=(x//60)/(10)rArr x = 60` Thus, 60 g of the solute must be added to 180 g of water so that the vapour pressure of water is lowered by 10%. |
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