25.0℃ CaCl2 sample 11.0 g water 125 g final temp

25.0℃ CaCl2 sample 11.0 g water 125 g final temp

 

 

Consider the dissolution of CaCl2:

CaCl2(s) ⟶ Ca^2+(aq) + 2Cl^-(aq) ... ΔH = –81.5 kJ

 

An 11.0 g sample of CaCl2 is dissolved in 125 g of water,

with both the substances at 25.0℃.

Calculate the final temperature of the solution

assuming no heat lost to the surroundings

and the solution has a specific heat capacity of 4.18 J/g•℃.

 

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CaCl2의 몰질량 = 110.98 g/mol

11.0 g / (110.98 g/mol) = 0.09912 mol CaCl2

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

(–81.5 kJ/mol) (0.09912 mol) = –8.07828 kJ = –8078.28 J

---> 11.0 g CaCl2가 방출한 열

---> 용액이 흡수한 열 = 8078.28 J

 

 

 

q = C m Δt

( 참고 https://ywpop.tistory.com/2897 )

 

Δt = q / C m

= (8078.28 J) / [(4.18 J/g•℃) (11.0 + 125 g)]

= 14.21℃

 

 

 

Δt = 최종온도 – 처음온도

25.0 + 14.21 = 39.2℃

 

 

 

답: 39.2℃

 

 

 

 

[참고] –81.5 kJ ≡ –81.5 kJ/mol

CaCl2(s) ⟶ Ca^2+(aq) + 2Cl^-(aq) ... ΔH = –81.5 kJ

1 mol CaCl2(s)가 용해될 때

–81.5 kJ만큼 방출되었으므로,

ΔH = –81.5 kJ ≡ ΔH = –81.5 kJ/mol

 

 

 

[키워드] 용액의 비열과 용액의 질량 기준문서