Calculate Molar Enthalpy of Hydrochloric Acid Solution

[Marzguerilla]

a laboratory technician adds 43.1 mL of concentrated, 11.6 mol/L hydrochloric acid to water to form 500 ml of dilute solution. the temperature of the solution changes from 19.2C to 21.8C. calculate the molar enthalpy of the equation.

so this is what is given
t=temperature (19.2-21.8= -2.6)
c=specific heat capacity. for water is 4.18 but i don't know if I am suppose water for this
m=mass I'm a bit lost in this one but i used 43.1ml/\H=mc/\t/n
this is what i tried

mHCl=43.1g t= -2.6 MHCl= 36.46 g/mol
43.1/36.46=1.18n mol
(43.1*4.18*-2.6)/ 1.182
=150.2

and i think the answer i got is wrong

 

[Cesium]

For your change in temperature remember that it is the final temperature minus the initial temperature. You've shown the reverse. If the problem doesn't give you any information about the specific heat of the solution then it is probably safe to use 4.18 J/g-C. In reality, the specific heat of this solution would be slightly different. Mass is measured in grams...you are using milliliters. To find the mass of the solution, use the density of water (again, another assumption...assume density of solution is equal to that of water- 1.0g/mL unless otherwise stated).

 

[Marzguerilla]

umm yea i know to convert mL to gram. I still believe I am getting the wrong answer. i don't know what's up with the mass but I am thinking its not 43.1. I am so lost now :(

 

[Marzguerilla]

does anyone know how to do this question? I am praying to god that there was a mistake in my chemistry book

 

[GCT]

a laboratory technician adds 43.1 mL of concentrated, 11.6 mol/L hydrochloric acid to water to form 500 ml of dilute solution. the temperature of the solution changes from 19.2C to 21.8C. calculate the molar enthalpy of the equation.

so this is what is given
t=temperature (19.2-21.8= -2.6)
c=specific heat capacity. for water is 4.18 but i don't know if I am suppose water for this
m=mass I'm a bit lost in this one but i used 43.1ml


/\H=mc/\t/n



this is what i tried

mHCl=43.1g t= -2.6 MHCl= 36.46 g/mol
43.1/36.46=1.18n mol
(43.1*4.18*-2.6)/ 1.182
=150.2

and i think the answer i got is wrong

It's too bad that they don't give you the initial temperature of the HCl solution, but for the sake of solving the problem with respect how your instructor probably wants you to do it, we should neglect this important aspect of calculating the molar enthalpy

You need to use the 500 mL to calculate the mass of water

Use this to calculate "q"

Find the moles of HCl and divide the "q" by this value

Arrive at the final units of J of KJ / moles HCl

 

[Gokul43201]

...
t=temperature (19.2-21.8= -2.6)
c=specific heat capacity. for water is 4.18 but i don't know if I am suppose water for this
m=mass I'm a bit lost in this one but i used 43.1ml


/\H=mc/\t/n

this is what i tried

m is the mass of the entire solution (assuming, as GCT pointed out, that the HCL is added at a temperature close to 19.2C). So that's the mass of about 543mL of liquid at a density of close to 1gm/mL.

 

[GCT]

m is the mass of the entire solution (assuming, as GCT pointed out, that the HCL is added at a temperature close to 19.2C). So that's the mass of about 543mL of liquid at a density of close to 1gm/mL.

The problem mentions "500 ml of dilute solution"; that is, 500 mL is the final volume right, unless I'm missing something here?

 

[Gokul43201]

The problem mentions "500 ml of dilute solution"; that is, 500 mL is the final volume right, unless I'm missing something here?

No, you're right. I didn't read carefully. The mass involved is that of 500mL of solution.