A student dissolved 2.50 grams of sodium hydroxide in 50.00 grams of water. The
ID: 551064 • Letter: A
Question
A student dissolved 2.50 grams of sodium hydroxide in 50.00 grams of water. The water’s temperature was initially 25.0oC, but after the sodium hydroxide dissolved, the temperature of the water was 29.5oC. Given that the specific heat of water is 4.18 J/goC complete the following calculations.
What was the change in temperature of the solution?
How much heat was lost or gained by the solution?
What is the H for the solvation of sodium hydroxide?
Calculate the H in kJ/g of NaOH.
Determine the H in kJ/mol.
Is the solvation endothermic or exothermic?
Write the complete thermo equation for the solvation of sodium hydroxide in water
Explanation / Answer
Ans mass of NaOH = 2.5 g
mass of water = 50 g
T1 = 25 celsius
T2 = 29.5 celsius
c = 4.18 J g-1 C-1
Change in temperature = T2 -T1 = 29.5 - 25.0 = 4.5 degree celsius
Heat lost = m c dT = 2.5 * 4.18 * 4.5 = 47.02 J
Solvation of NaOH in water is an exofhermic reaction because heat is released during the course of reaction.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.