Gc h

[LSDP]

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The ΔH for the solution process when solid sodium hydroxide dissolves in water is 44.4 kJ/mol. When a 13.9-g sample of NaOH dissolves in 250.0 g of water in a coffee-cup calorimeter, the temperature increases from 23.0°C to ________°C. Assume that the solution has the same specific heat as liquid water, i.e., 4.18 J/g-K.

A) 35.2°C
B) 24.0°C
C) 37.8°C
D) 37.0°C
E) 40.2°C

difficulty 4/5

What is the answer?
I hate problem like this, I don't know what mass to use.
thanks in advance. Good practice if you are going to take the DAT soon.

 

[Kingsta2426]

The ΔH for the solution process when solid sodium hydroxide dissolves in water is 44.4 kJ/mol. When a 13.9-g sample of NaOH dissolves in 250.0 g of water in a coffee-cup calorimeter, the temperature increases from 23.0°C to ________°C. Assume that the solution has the same specific heat as liquid water, i.e., 4.18 J/g-K.

A) 35.2°C
B) 24.0°C
C) 37.8°C
D) 37.0°C
E) 40.2°C

difficulty 4/5

What is the answer?
I hate problem like this, I don't know what mass to use.
thanks in advance. Good practice if you are going to take the DAT soon.

I got 40.2 by doing (44400 ) / (263.9 x 4.184) idk if its correct though bc I have never done a problem asking for delta T like this.

 

[LSDP]

sorry BROW it's not E. you forgot about the temperature change. 40-23=17
When you plug 40.2 back to the equation you don't get 44 44.4 kJ/mol.
263.9(4.184)(40.5-23) = 18688 J.

This obviously lack of knowledge, not arithmetic issue. I can't figure it out.

the problem is that the other side is 44.4kJ/mol. How do I convert 44.4kJ/mol to mol?


why is the answer D.

 

[Miley13]

Dont let the units trick you! You need delta h in joules for the units to all cancel. Do this by:

(44,400 J / mole NaOh) x (1 mole NaOH / 40.0 g NaOH) = 1110 J / g NaOH

(1110 J / g NaOH) x 13.9 g NaOH = 15,400 J

15,400 J is the heat released by dissolving 13.9 g NaOH in water. The heat released by dissolving is absorbed by the water.


Delta H = (mass H2O)(specific heat H2O)(Tf - Ti)
15,400 J = (250.0 g)(4.184 J/g K)(Tf - 23C)
39458 = 1046 x Tf
Tf = 37.7 C

 

[jhc83]

where did you see this problem?? topscore or achiever??

 

[LSDP]

Omg thank you. You know your materials well.

This is a good question. I got this question from brown chemistry book 11e. I'm trolling for questions.