Ice at 0.0 degree C is used to cool water. What is the minimum mass of ice requi
ID: 531288 • Letter: I
Question
Ice at 0.0 degree C is used to cool water. What is the minimum mass of ice required to cool 323 g of water from 30.5 degree C to 4.0 degree C? (Heat of fraction = 333 J/g; specific heat capacities ice = 2.06 J/g K. liquid water = 4.184 J/g K) a) 108 g b) 125 g c) 325 g d) 605 g e) 1.75 times 10^- Salt A and salt B were dissolved separately in 100 mL beakers of water. The water temperature were measured and recorded as shown below: Salt A: initial water temp. 25.1 degree C final water temp. 30.2 degree C Salt B: initial water temp. 25.1 degree C final water temp 20.0 degree C Which statement is a correct interpretation of these data? a) Dissolving salt A was endothermic b) Dissolving salt B was endothermic c) Dissolving salt B was exothermic d) Dissolving both salt A and salt B was endothermic. e) Dissolving salt A was exothermic and dissolving salt B was endothermic. The pressure of N_3 in a 20.0 L flask in 0.603 atm at 62 degree C, What mass of N_2 is in the flask? (K = 0.03206 L atm/mol K) a) 0.439 g b) 6.15 g c) 12.3 g d) 63.9 g e) 66.4 gExplanation / Answer
26. Solution A containing salt A:
After adding the salt to water, the temperature of the mixture is raising.
This indicates that it is an exothermic reaction.
Solution B containing salt B:
After adding the salt to water, the temperature of the mixture is decreasing.
This indicates that it is an endothermic reaction.
Hence option e is correct.
27. The mass of N2 gas can be calculated by using ideal gas equation.
PV = (mass of the gas/its molecular mass)x R.T
Where P =pressure of the gas = 0.603atm
V =volume of the gas = 20L
R= universal gas constant = 0.08206L.atm.mol-1.K-1
T = temperatiure of the gas = 620C = (62+273)K = 335K
We know the molecular mass of N2 gas = 28g/mol
by rearranging the ideal gas equation, we get mass of the gas:
mass of the gas = PV. Molecular mass of the gas/RT
By substituting the given values, we get:
mass of the gas = 0.603atm. 20L .28g/mol / 0.08206L.atm.mol-1.K-1 . 335K
= 12.3g
mass of the gas =12.3g
25. The amount of heat energy lost by water = mass of water. deltaT . specific heat capacity of water
= 325g (30.5-4)K .4.184J/g.K
= 325g . 26.5K . 4.184 J/g.K
= 36.034kJ
This heat energy lost by water is due to ice at 00C.
Hence 36.034kJ = mass of ice . latest heat of fusion of ice
By rearranging
=> mass of ice = 36.034kJ/latest heat of fusion of ice
= 36.034kJ/0.333 kJ/g
= 108.2g
hence 108.2g of ice is required to cool 325g of water from 30.50C to 4.00C.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.