NaNO3 KF mixture 10.0 g water 100.0 g 2.22℃

NaNO3 KF mixture 10.0 g water 100.0 g 2.22℃

 

 

Imagine that you dissolve 10.0 g of a mixture of NaNO3 and KF

in 100.0 g of water and find that the temperature rises by 2.22℃.

Using the following data,

calculate the mass of each compound in the original mixture.

Assume that the specific heat of the solution is 4.18 J/g•℃.

NaNO3(s) → NaNO3(aq) ... ΔH = +20.4 kJ/mol

KF(s) → KF(aq) ... ΔH = –17.7 kJ/mol

 

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NaNO3의 몰질량 = 84.99 g/mol 이므로,

ΔH = (+20.4 kJ/mol) / (84.99 g/mol) = +0.240028 kJ/g

 

KF의 몰질량 = 58.10 g/mol 이므로,

ΔH = (–17.7 kJ/mol) / (58.10 g/mol) = –0.304647 kJ/g

 

 

 

q = C m Δt

( 참고 https://ywpop.tistory.com/2897 )

 

= (4.18) (10.0 + 100.0) (2.22)

= 1020.756 J

= 1.020756 kJ

---> 용액이 흡수한 열

---> 반응 계가 방출한 열 = –1.020756 kJ

 

 

 

NaNO3의 질량 = x 라 두면,

KF의 질량 = (10.0 – x)

 

 

 

–1.020756 kJ = (+0.240028 kJ/g) (x g) + (–0.304647 kJ/g) (10.0 – x g)

 

–1.020756 = 0.240028x – 3.04647 + 0.304647x

 

0.544675x = 2.025714

 

x = 2.025714 / 0.544675 = 3.719 g

---> NaNO3의 질량

 

10.0 – 3.719 = 6.281 g KF

 

 

 

답: 3.72 g NaNO3, 6.28 g KF