Calculating molar concentration and molar enthalapy...HELP PLZ!!

A student determined the enthalpy of NaOH by dissolving 4.00 g of NaOH(s) in 180 mL of pure water (density 0.999 g/mL) at an initial temperature of 19.5 °C. Temperature-time data collected after mixing was extrapolated back to the time of mixing to obtain a temperature change, ΔT = +5.5 °C. The density of the final solution was 1.013 g/ml at 25 °C, and the solution heat capacity was 4.08 J/(g °C). In a separate experiment, the calorimeter constant was found to be 21 J/°C.

1.) What is the molar concentration of the final NaOH solution at 25°C?

2.) What is the molar enthalapy of dissolution of NaOH in water?

I really need help on these two questions. Thanks in advance!

Hello

you have 184 g solvent plus solute with a density of 1.013 g/mL at 25°C.
1.013 g = 1mL
1 g = 1/1.013 mL
184 g = 184/1.013 mL = 181.638 mL total volume
Now you have 4 g NaOH = 0.100 moles in 181.638 mL
= 0.1*1000/181.638 moles in 1000 mL = 0.55054 moles in 1 L at 25°C = 0.5505 M NaOH
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The heat given off by the water/NaOH solution is
cm*∆T + 21J*∆T
4.08*184*5.5 + 21*5.5 = 4244.46 J
and since 4 g = 0.100 moles the molar enthalpy = - 42444.6 J/mol

Regards