Question

1.0 g of sodium hydroxide, NaOH, was added to 99.0 g of water. The temperature of the solution increased from 18.0 ºC to 20.5 ºC. The specific heat capacity of the solution is
4.18 J g–1 K–1.
Which expression gives the heat evolved in kJ mol–1?

Answer 1

Using \[Q=mc \Delta T\]
Q=(99.0)(4.18)(2.5)
Q=1034.55J

1.0 g of sodium hydroxide=0.0435mol Na
Then use \[\Delta H={Q \over n}\]
And get your answer
\[\Delta H =\frac{1034.55}{0.0435}\]

Answer 2

oh I didn't see that theres a mistake
\[1gNaOH=\frac{1molNaOH}{40gNaOH} \times \frac{1molNa}{1molNaOH} =0.25molNa\]
Then use \[\Delta H={Q \over n}\]
And get your answer
\[\Delta H =\frac{1034.55}{0.25}\]
Thanks for catching @Callisto :)