A 14.1g sample of NaOH is dissolved in 200.0g of water in a coffee-cup colorimeter. The temperature increases from 20.0C to —-. Specific heat of liquid water is 4.18 J/g-K and deltaH for the dissolution of sodium hydroxide in water is 44.4kj/mol.

The temperature finally increased from 20 °C to 38.7 °C

Explanation:

This a problem about calorimetry. Formula to solve it is:

Q = m . C . ΔT

Where ΔT is Final T° - Initial T°

m is the mass

and C, specific heat.

In this situation we are dissolving NaOH where 1 mol requires 44.4 kJ of heat. As we do not have 1 mol, we need to determine the heat that is released.

We convert the mass to moles

14.1 g . 1 mol / 40g = 0.3525 mol

We determine the heat

0.3525 mol . 44400 J/mol = 15651 Joules of heat.

Now we can replace data in the calorimetry formula:

15651 J = 200 g . 4.18 J/g°C (Final T° - 20°C)

15651 J / (200 g . 4.18 J/g°C) = Final T° - 20°C

18.7°C + 20°C = Final T° → 38.7°C

im pretty sure its b