Answer: The boiling point of a 3.70 m solution of phenol in benzene is [tex]89.5^0C[/tex]
Explanation:
Elevation in boiling point:
[tex]\Delta T_b=i\times k_b\times m[/tex]
where,
[tex]\Delta T_b[/tex] = change in boiling point
i= vant hoff factor = 1 (for benzene which is a non electrolyte )
[tex]k_b[/tex] = boiling point constant = [tex]2.53^0C/kgmol[/tex]
m = molality = 3.70
[tex]T_{solution}-T_{solvent}=i\times k_b\times m[/tex]
[tex]T_{solution}-80.1^0C=1\times 2.53\times 3.70[/tex]
[tex]T_{solution}=89.5^0C[/tex]
Thus the boiling point of a 3.70 m solution of phenol in benzene is [tex]89.5^0C[/tex]
Answer:
First Question:
C. ΔTb = Kbm
Second Question:
C. 9.36° C
Third Question:
89.5
Explanation:
Got it right.
