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InterviewSolution
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The empirical formula of a non-electrolyte is `CH_(2)O`. A solution containing 3 g `L^(-1)` of the compound exerts the same osmotic pressure as that of 0.05 M glucose solution. The molecular formula of the compound is :A. `C_(2)H_(4)O_(2)`B. `C_(3)H_(6)O_(3)`C. `C_(5)H_(10)O_(5)`D. `C_(4)H_(8)O_(4)` |
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Answer» Correct Answer - D For isotonic solution : `(w_(1))/(m_(1)V_(1))=(w_(2))/(m_(2)V_(2))` `w_(1` = mass of glucose `= 0.05xx180g = 9g` `m_(1)` = molecular mass of glucose = 180 g Assuming `V_(1)=V_(2)=1L` `w_(2)` = mass of compound = 6g `m_(2)` = molecular mass of compound = ? `therefore (9)/(180)=(6)/(x)rArr x=120g` `therefore ("Molecularmass")/("Empiricalmass")=n rArr=(120)/(30)=4` `therefore` Molecular formula `= C_(4)H_(8)O_(4)`. |
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