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The mass of a non-volatile solute of molar mass `40g" "mol^(-1)` that should be dissolved in 114 g of octane to lower its vapour pressure by 20% isA. `11.4g`B. `9.8g`C. `12.8g`D. 10g |
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Answer» Correct Answer - D `"Given, "p^(@)=100,p=100-20=80`, `M_(2)=40" gmol"^(-1),M_(1)("octane")=C_(8)H_(18)=114" gmol"^(-1)` `w_(2)=?,w_(1)("octane")=114g` `(p^(@)-p)/(p)=(w_(2))/(M_(2))xx(M_(1))/(w_(1)` `(100-80)/(80)=(w_(2))/(40)xx(114)/(114)` `rArrw_(2)=10g` |
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