Calculating approximate heat of a solution?

[Mxrkko]

Hey guys, not sure where to start with this question?

When 2.0g of solid sodium hydroxide is dissolved in water to form 100g of solution, the temperature increases by 5.0 degrees celsius. What is the approximate heat of solution, in kilojoules mol/L?

[Mxrkko]

Forgot to mention that the possible answers are
a) 42, exothermic
b) 42, endothermic
c) 0.8, exothermic
d) 0.8, endothermic

[RuiAce]

Are you absolutely certain that the units are kJ mol/L?

Because if it were the correct units kJ /mol then I have a method to show why the answer is a.


(Obviously it's exothermic because the temperature increased.)

[RuiAce]

mol^-L?

Either way, the question is clearly flawed then as approximate heat of solution is essentially enthalpy change (aka molar heat of reaction) which is meausred in kJ mol^-1

q = m c ΔT
q = 0.1kg * 4.18*10³ J kg^-1 K^-1 * 5 K
q = 2090 J = 2.090 kJ

n_NaOH = m/MM = 2.0/(22.99+16.00+1.008) = 0.05000250013... mol

ΔH = -q/n
= -2.090/0.05000250013...
= 41.79791 kJ mol^-1

[Mxrkko]

Thank you so much! It's going to take some time to fully understand this, as we haven't even started the energy topic yet, but it's good to have an early start! Thanks again

[RuiAce]

If you just started, here's some reassurance:

This is VERY hard when you just start. It's one of the easiest things to get lost in chemistry