Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Experimental Determination of the Heat of Solution for KNO3, Exercises of Chemistry

The data and calculations for determining the heat of solution for potassium nitrate (KNO3) based on three experiments where different amounts of KNO3 were added to water and the resulting change in temperature was measured. The specific heat of water was used to calculate the heat of solution per mole of KNO3.

Typology: Exercises

2021/2022

Uploaded on 09/12/2022

thecoral
thecoral 🇺🇸

4.4

(29)

401 documents

1 / 8

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
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) oC = -2.1oC
q = mcΔT= 101.15 g x 4.184 j/goC x -2.1oC= -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 KNO3 so it is +
heat of solution per mole(J)= 888.74 J/0 .0255 mol= 34829.82257 J/mol= + 35
K J/mol
pf3
pf4
pf5
pf8

Partial preview of the text

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