Please help! I have tried this problem so many ways and it\'s wrong every time!
ID: 916316 • Letter: P
Question
Please help! I have tried this problem so many ways and it's wrong every time! Thank you! previous | 29 of 60 next art A A calorimeter contains 24.0 mL of water at 14.5 C. When 2.50 g of X (a substance with a molar mass of 73.0 g/mol ) is added, it dissolves via the reaction x(s) + H2O(l)-x(aq) and the temperature of the solution increases to 28.5 C Calculate the enthalpy change, AH, for this reaction per mole of X. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/g·°C)], that density of water is 1.00 g/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings Express the change in enthalpy in kilojoules per mole to three significant figures AH40.88 kJ/mol Submlt Hints My Answers Give Up Review Part Incorrect; Try Again; 3 attempts remaining In the calculation of g the total mass of reactant and water is needed. You may have used just the mass of the water, Part B Consider the reaction C12H22011 (s) t 120, (g) 12CO2 (g) + 11H20(1) in which 10.0 g of sucrose. C12H22 01. was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/ C. The temperature increase inside the calorimeter was found to be 22.0 C. Calculate theExplanation / Answer
Answer – Given, volume of water = 24.0 mL , mass of X = 2.50 g
Molar mass of X = 73.0 g/mol , initial temp , ti =14.5 oC , final temp , tf = 28.5 oC
Specific heat of water = 4.184 J/goC, density of solution = 1.0 g/mL
So, mass of solution = 24.0 * 1.0 g/mL = 24.0 g + 2.25 g = 26.5 g
We know, the formula for calculating the heat
Heat, q = m*C*t
= 26.5 g * 4.184 J/goC *( 28.5-14.5)oC
= 1552.3 J
= 1.552 kJ
There is no heat loss, means heat loss from the X = heat gain by the solution
So, heat of X, q = -1.552 kJ
We also, know, H = q
So, H =-1.552 kJ
Moles of X = 2.50 g / 73.0 g.mol-1
= 0.0342 moles
So, H =-1.552 kJ /0.0342 moles
= -45.3 kJ/mol