Answer: The molar mass of the hydrocarbon is 37 g/mol
Explanation:
Elevation in boiling point is given by:
[tex]\Delta T_b=i\times K_b\times m[/tex]
[tex]\Delta T_b=T_b-T_b^0=(56.50-55.95)^0C=0.55^0C[/tex] = Elevation in boiling point
i= vant hoff factor =1 (for non electrolyte )
[tex]K_b[/tex] = boiling point constant = [tex]0.512^0C/m[/tex]
m= molality
[tex]\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (acetone) = 95 g = 0.095 kg
Molar mass of solute = M g/mol
Mass of solute = 3.75 g
[tex]0.55^0C=1\times 0.512\times \frac{3.75g}{Mg/mol\times 0.095kg}[/tex]
[tex]M=37g/mol[/tex]
Thus the molar mass of the hydrocarbon is 37 g/mol
Given:
[tex]= 56.50-55.95[/tex]
[tex]= 0.55^{\circ} C[/tex]
As we know,
→ [tex]\Delta T_b = i\times K_b\times m[/tex]
or,
→ [tex]\Delta T_b = i\times K_b\times \frac{Mass \ of \ solute}{Molar \ mass \ of \ solute\times Weight \ of \ solvent}[/tex]
By putting the values,
→ [tex]0.55 = 1\times 0.512\times \frac{3.75}{M\times 0.095}[/tex]
[tex]M = 35 \ g/mol[/tex]
Thus the above answer is correct.
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