A 5.0-g Sample of KBr at 25.0 degrees celsius dissolves in 25.0 degrees celsius 25.0 mL of water also at 25.0 degrees celsius. The final equilibrium temperature of the resulting soltution is 18.1 degrees celsius. What is the enthalpy of of solution in kilojoules per mole of KBr?

Try reading your question. It needs to be cleaned up. It makes no sense as is.

-q = dH ( dH = change in enthalpy Kj/mol)


Use q=mc(Tf-Ti) and then convert to deltaH

q=mc(Tf-Ti)

q =?
c= 4.814 J/g*C use specific heat of water since its in aqueous sol and we are not told differently

m= total mass= mass KBr + mass water
mass KBr= 5.0 g
*mass of water from density =1g/mL so
(25.0ml) (1g/ml)=25.0g water
m = 25.0g water + 5.0 g KBr = 30.0g

Tf-Ti= 18.1 - 25.0 = -6.9C

q= (30g) (4.184 J/gC)(-6.9C)
q= -1443 J

-q = delta H kJ/mol KBr

dH= (-q Joules)( 1 kJ/1000 Joules) / mol KBr

*mol KBr = (5.0 g KBr)(1mol KBr) /119g KBr)= 0.042 mol KBr

dH = -(-1443 J) (1kJ/1000J)/ 0.042 mol KBr)

dH= 34.4 kJ/mol KBr

To find the enthalpy of solution, we can use the equation:

ΔH_solution = q_solution / n

Where:
ΔH_solution is the enthalpy of solution in kilojoules per mole (kJ/mol)
q_solution is the heat released or absorbed during the solution process in joules (J)
n is the number of moles of solute (KBr)

First, let's calculate the heat released or absorbed during the solution process.

q_solution = mcΔT

Where:
m is the mass of water (in kg)
c is the specific heat capacity of water (4.18 J/g·°C)
ΔT is the change in temperature (in °C)

We are given:
Mass of KBr = 5.0 g
Initial temperature of water = 25.0 °C
Final equilibrium temperature = 18.1 °C
Volume of water = 25.0 mL = 25.0 g

Converting the mass of water to kg:
Mass of water = 25.0 g = 0.025 kg

Calculating the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 18.1 °C - 25.0 °C
ΔT = -6.9 °C

Now, let's calculate the heat released or absorbed during the solution process:

q_solution = mcΔT
q_solution = (0.025 kg)(4.18 J/g·°C)(-6.9 °C)
q_solution = -0.72275 J

Since we need the enthalpy of solution in kilojoules per mole (kJ/mol), we need to convert the heat change to kilojoules and calculate the number of moles of KBr.

1 J = 0.001 kJ
1 g = 0.001 kg
Molar mass of KBr = 119 g/mol

Converting heat change to kilojoules:
q_solution = -0.72275 J = -0.72275 x 0.001 kJ
q_solution = -0.00072275 kJ

Calculating the number of moles of KBr:
Number of moles = mass / molar mass
Number of moles = 5.0 g / 119 g/mol
Number of moles = 0.04202 mol

Now, let's calculate the enthalpy of solution:

ΔH_solution = q_solution / n
ΔH_solution = (-0.00072275 kJ) / 0.04202 mol
ΔH_solution = -17.2068 kJ/mol

Therefore, the enthalpy of solution for KBr is approximately -17.21 kJ/mol.

To calculate the enthalpy of solution, we can use the equation:

ΔH_solution = q_solution / n

Where:
ΔH_solution = enthalpy of solution
q_solution = heat exchanged during the solution process
n = amount of solute (in moles)

First, we need to calculate the heat exchanged during the solution process, q_solution. We can use the equation:

q_solution = m_cΔT

Where:
m = mass of the water (in grams)
c = specific heat capacity of water (4.18 J/g·°C)
ΔT = change in temperature (final temperature - initial temperature)

Given:
Mass of water (m) = 25.0 mL
Temperature change (ΔT) = 25.0°C - 18.1°C = 6.9°C

To convert the mass of water from mL to grams, we need to consider the density of water:

Density of water = 1.0 g/mL

Therefore, the mass of water is 25.0 g.

Now we can calculate q_solution:

q_solution = (25.0 g) * (4.18 J/g·°C) * (6.9°C)

Next, we need to calculate the amount of solute, n, using the sample size and molar mass of KBr.

The molar mass of KBr is:
M(K) = 39.10 g/mol
M(Br) = 79.90 g/mol

Molar mass of KBr = M(K) + M(Br) = 39.10 g/mol + 79.90 g/mol = 119.00 g/mol

Given:
Sample size = 5.0 g

n = (5.0 g) / (119.00 g/mol)

Now we can calculate the enthalpy of solution (ΔH_solution):

ΔH_solution = q_solution / n

Finally, convert the enthalpy of solution to kilojoules per mole by dividing by 1000.

I will now perform the calculations and provide you with the final answer.