Question

10.0 g of ammonium nitrate (?Hsoln = 25.7 kJ/mol, molar mass = 80.0 g/mol) dissolves in 100.0 g of water.

What is the change in temperature of the solution?

Assume the specific heat capacity of the solution is 4.2 J

Answer 1

Step-by-step explanation:

The given data is as follows.

Molar mass of ammonium nitrate = 80.0 g/mol

So, we will calculate the number of moles of ammonium nitrate as follows.

No. of moles =
`\frac{\text{given mass}}{\text{molar mass}}`

=
`(10.0 g)/(80.0 g/mol)`

= 0.125 mol

Heat released due to solution of ammonium nitrate =
`\Delta H * \text{no. of moles}`

=
`25.7 kJ/mol * 0.125  mol`

= 3.2125 KJ

= 3212.5 J (as 1 kJ = 1000 J)

Therefore, calculate the total mass of solution as follows.

mass of solution(m) = (10.0 + 100.0 ) g

= 110.0 g

Hence, heat released will be calculated as follows.

Q =
`m * C * \Delta T`

3212.5 J =
`110.0 * 4.2 J * \Delta T`

`\Delta T = 6.95^(o)C`

Thus, we can conclude that the change in temperature of the solution is
`6.95^(o)C`.

Answer 2

The change in temperature = 7.65 °C

Step-by-step explanation:

Step 1: Data given

10.0 grams of ammonium nitrate dissolves in 100.0 grams of water

Hsoln = 25.7 kJ/mol

Molar mass = 80.04 g/mol

Heat capacity of the solution = 4.2 J

Step 2: Calculate moles ammonium nitrate

Moles = mass / molar mass

Moles = 10.0 grams / 80.04 g/mol

Moles =0.125 moles

Step 3: Calculate q

q = 25.7 kJ/mol * 0.125 moles

q = 3.2125 kJ = 3212.5 J

Step 4: Calculate change in temperature

q = m*c*ΔT

3212.5 J = 100g *4.2 J * ΔT

ΔT= 7.65

The change in temperature = 7.65 °C