C12H10 5.50 g benzene 100.0 g 0.903℃ 6.30 g 150.0 g 0.597℃

C12H10 5.50 g benzene 100.0 g 0.903℃ 6.30 g 150.0 g 0.597℃

 

 

(a) When 5.50 g of biphenyl (C12H10)

is dissolved in 100.0 g of benzene (C6H6),

the boiling point increases by 0.903℃.

Calculate the Kb for benzene.

 

(b) If 6.30 g of an unknown were added to 150.0 g of benzene,

the boiling point of the solution increases by 0.597℃.

What is the molar mass of the unknown substance?

 

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5.50 g / (154.21 g/mol) = 0.0357 mol C12H10 (용질)

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

몰랄농도 = 용질 mol수 / 용매 kg수

= 0.0357 mol / (100.0/1000 kg)

= 0.357 mol/kg

= 0.357 m

 

 

 

ΔTb = Kb × m

( 참고 https://ywpop.tistory.com/1920 )

 

Kb = ΔTb / m

= 0.903 / 0.357

= 2.53 ℃/m

---> (a) 답

 

 

 

 

ΔTb = Kb × m

 

m = ΔTb / Kb

= 0.597 / 2.53

= 0.236 m

 

 

 

0.236 m = ? mol / (150.0/1000 kg)

 

? = 0.236 × (150.0/1000)

= 0.0354 mol 용질

 

 

 

몰질량 = 질량(g) / 몰수(mol)

= 6.30 g / 0.0354 mol

= 178 g/mol

---> (b) 답