An organic compound conatins 40% C and 6.7% H.The boiling point of a 5% aqueous solution of this compound is found to be 373.15 K.Find the molecular formula mass of the compound . $K_{b}$ of water is 0.52.

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Solution

given C%=40
H%=6.7
O%=100-(40+6.7)=53.3
To calculate empirical formula
Atom % %Aw % Aw/Minimum Value
C 40 $\frac{40}{12} = 3.33$ 3.33/3.33=1
H 6.7 $\frac{6.7}{1} = 6.7$ 6.7/3.33=2
O 53.3 $\frac{53.3}{16} = 3.33$ 3.33/3.33=1
$∴$ Empirical formula of compound = $C H_{2} O$
molecular weight of E.formula compound =30
Now,given $Δ T_{b} = 0.15 K, K_{b} = 0.52$
Composition of solution =5%
Therefore, weight of solute $W_{2}$ =5 g
weight of solvent $W_{1}$ =100-5=95 g
We know
$Δ T_{b} = K_{b} m = K_{b} \times \frac{W_{2} \times 1000}{M w_{2} \times W_{1}}$
Substituting all values,we get
$0.15 = \frac{0.52 \times 5 \times 1000}{M w_{2} \times 95} \Rightarrow M w_{2} = 182.45$
$∴ n = \frac{C a l c u l a t e d M w}{E m p i r i c a l M w}$
$= \frac{182.45}{30} \approx 6$
$∴$ Molecular formula is ${(C H_{2} O)}_{6} = C_{6} H_{12} O_{6}$
and mol wt is 180.

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