Download Experimental Determination of the Heat of Solution for KNO3 and more Exercises Chemistry in PDF only on Docsity!
A 2.58-g sample of KNO3 (f.w. = 101.11 u) was added to 98.57 g of water in a
coffee-cup
calorimeter. The initial temperature of the water was 22.5 oC and the
temperature of the
solution after mixing was 20.4 oC. On the basis of this experiment, what is the
heat of
solution per mole of KNO3 [f.w. = 101.11 u]? The specific heat of water is
4.184 J/goC.
Total mass of the solution =g of KNO3 + g of water
= (2.58+ 98.57) g
= 101.15 g
Change in temperature =(20.4 – 22.5)
o C = -2.
o C
q = mcΔT= 101.15 g x 4.184 j/g
o
C x -2.
o
C= -8 8.74 J
Moles of KNO3= 2.58/101.11= 0.0255 mol
The heat that is being lost by the water is being gained by KNO 3 so it is +
heat of solution per mole(J)= 888.74 J/0 .0255 mol= 34829.82257 J/mol= + 35
K J/mol
A 3.53-g sample of KNO3 (f.w. = 101.11 u) was added to 97.77g of water in a
coffee-cup
calorimeter. The initial temperature of the water was 22.5 oC and the
temperature of the
solution after mixing was 20.4 oC. On the basis of this experiment, what is the
heat of
solution per mole of KNO3 [f.w. = 101.11 u]? The specific heat of water is
4.184 J/goC.
Total mass of the solution =g of KNO3 + g of water
= (3.53+ 97.77) g
= 101.30 g
Change in temperature =(20.4 – 22.5)
o
C = -2.
o
C
q = mcΔT= 101.30 g x 4.184 j/g
o C x -2.
o C= -890.06J
Moles of KNO3= 3.53/101.11= 0.0349 mol
The heat that is being lost by the water is being gained by KNO 3 so it is +
heat of solution per mole(J)= 890.06 J/ 0.0349 mol =25494.10798 J/ mol = +
25KJ/mol
A 5.58-g sample of KNO3 (f.w. = 101.11 u) was added to 95.55g o f water in a
coffee-cup
calorimeter. The initial temperature of the water was 22.5 oC and the
temperature of the
solution after mixing was 20.4 oC. On the basis of this experiment, what is the
heat of
solution per mole of KNO3 [f.w. = 101.11 u]? The specific heat of water is
4.184 J/goC.
Total mass of the solution =g of KNO3 + g of water
= (5.58+ 95.55) g
= 101.13 g
Change in temperature =(20.4 – 22.5)
o
C = -2.
o
C
q = mcΔT= 101.13 g x 4.184 j/g
o C x -2.
o C= -888.57 J
Moles of KNO3= 5.58/101.11= 0.0552 mol
The heat that is being lost by the water is being gained by KNO 3 so it is +
heat of solution per mole(J)=888.57 J/.0552mol = 16100.92731 = 16KJ/mol
A 20.00-mL sample of sodium chloride solution was titrated with 0.4000 M
AgNO
solution, requiring 28.62 mL to reach the equivalence point.
The titration reaction equation is
Cl
( aq ) + AgNO3( aq ) --> AgCl( s ) + NO
( aq )
A. What was the concentration of Cl
–– ion in the original sample
The reaction is 1:1 for AgNO3 and Cl-
Moles of AgNO3 = volume of AgNO3 in L x molarity of AgNO
= 0.02862L x 0.4000 M = 0.01145 moles
Concentration of Cl- = moles of Cl-/ volume of Cl- in L
= 0.01145 moles/0 .02000 L= 0.5724 M
B. How many grams of precipitate were formed?
(f.w. Ag Cl = 143.32 u).
AgNO3 and AgCl are 1:1 molar ratio
Moles of AgCl = 0.01145 moles
g of AgCl= 0.01145 moles * 143.32 = 1.641 grams
A 27.00-mL sample of sodium chloride solution was titrated with 0.5000 M
AgNO
solution, requiring 26.62 mL to reach the equivalence point.
The titration reaction equation is
Cl
( aq ) + AgNO3( aq ) --> AgCl( s ) + NO
( aq )
A. What was the concentration of Cl
–– ion in the original sample
The reaction is 1:1 for AgNO3 and Cl-
Moles of AgNO3 = volume of AgNO3 in L x molarity of AgNO
=0 .02662L x0 .5000 M = 0.01331 moles
Concentration of Cl- = moles of Cl-/ volume of Cl- in L
= 0.01331 moles/ .02700 L= 0.4930 M
B. How many grams of precipitate were formed?
(f.w. Ag Cl = 143.32 u).
AgNO3 and AgCl are 1:1 molar ratio
Moles of AgCl = 0.01331 moles
g of AgCl= 0.01331 moles x 143.32 = 1.908 grams
A 29.00-mL sample of sodium chloride solution was titrated with 0.5500 M
AgNO
solution, requiring 25.62 mL to reach the equivalence point.
The titration reaction equation is
Cl
( aq ) + AgNO3( aq ) --> AgCl( s ) + NO
( aq )
A. What was the concentration of Cl
–– ion in the original sample
The reaction is 1:1 for AgNO3 and Cl-
Moles of AgNO3 = volume of AgNO3 in L x molarity of AgNO
= 0.02562L x 0.5500 M = 0.01409 moles
Concentration of Cl
= moles of Cl
/ volume of Cl- in L
= 0.01409 moles/ 0.02900 L= 0.4859 M
B. How many grams of precipitate were formed?
(f.w. Ag Cl = 143.32 u).
AgNO3 and AgCl are 1:1 molar ratio
Moles of AgCl = 0.01409 moles
g of AgCl= 0.01409 moles * 143.32 = 2.020 grams
