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일반화학/[13장] 용액의 성질

boiling point 102.36℃ water 250 g 180.9 g mol^-1 glucose gram

by 영원파란 2022. 10. 1.
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boiling point 102.36℃ water 250 g 180.9 g mol^-1 glucose gram

 

 

How many grams of glucose (molar mass = 180.9 g/mol)

must be dissolved in 250 g of water

to raise the boiling point to 102.36℃?

 

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끓는점오름, ΔTb

ΔTb = Kb × m

( 참고 https://ywpop.tistory.com/1920 )

 

 

 

m = ΔTb / Kb

= 2.36 / 0.51

= 4.62745 mol/kg

 

 

 

몰랄농도 = 용질 mol수 / 용매 kg수 이므로,

4.62745 mol/kg = ? mol / 0.250 kg

 

? = 4.62745 × 0.250

= 1.1568625 mol

 

 

 

1.1568625 mol × (180.9 g/mol)

= 209.3 g

 

 

 

답: 209.3 g

 

 

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