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Calculate the mass of a non - volatile solute (molar mass 40 g) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%. |
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Answer» <P> Solution :Let the vapour pressure of pure OCTANE be `p_(1)^(0)``(80)/(100)p_(1)^(0)=0.8 p_(1)^(0)` Molar mass of solute, `M_(2)=40 g mol^(-1)` Mass of octane, `w_(1)=114 g` Molar mass of octane, `(C_(3)H_(18))`, `= M_(1)=8xx12+18xx1=114 g mol^(-1)` Applying the reaction, `(p_(1)^(0)-p_(1))/(p_(1)^(0))=(w_(2)xxM_(1))/(M_(2)xx w_(1))` `therefore (p_(1)^(0)-0.8 p_(1)^(0))/(p_(1)^(0))=(w_(2)xx114)/(40xx114)` `therefore (0.2 p_(1)^(0))/(p_(1)^(0))=(w_(2))/(40)` `therefore w_(2)=8 g` |
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