A student added 5.350 g of ammonium chloride to 100.00 cm3 of water. the initial temperature of the water was 25.55℃ but it decr

Question

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A student added 5.350 g of ammonium chloride to 100.00 cm3 of water. the initial temperature of the water was 25.55℃ but it decr

eased to 21.79℃. calculate the enthalpy change that would occur when 1 mol of the solute is added to 1.0000 dm3 of water.

Chemistry

2 answers:

eduard [2.7K] · Answer 1 · 2026-02-27T17:44:45+03:00

Answer:

When 1 mole of the solute is added to 1.0000 $dm^3$ of water, the enthalpy change is 157.168 kJ/mol.

Explanation:

Mass of ammonium chloride used = 5.350 g

Moles of ammonium chloride = $\frac{5.350 g}{53.5 g/mol}=0.1 mol$

Let Q be the heat absorbed by 0.1 mole of solute and Q' be the heat lost by water.

Thus, Q = -Q'

Water volume, V = $100.00 cm^3=100.00 mL$

Water mass = m

Density of water, d = 1 g/mL

$m=Density\times volume =d\times V=1 g/ml\times 100.00 mL=100.00 g$

The temperature change in the water = ΔT = 21.79°C - 25.55°C = -3.76°C

Specific heat of water = c = 4.18 J/g°C

$Q'=mc\Delta T$

$Q'=100.00 g\times 4.18J/g^oC\times (-3.76^oC)=-1571.68 J$

Q= -Q' = -(-1571.68 J) = 1571.68 J

Thus, 0.1 mole of solute absorbed 1571.68 Joules of energy from $100.00 cm^3$ of water.

When 1 mole of solute is dissolved in $100.00 cm^3$ of water:

$\frac{1571.68 J}{0.1}=15716.8 Joule$

$1 dm^3=1000 cm^3$

Thermal energy absorbed when dissolving 1 mole of solute in $100.00 cm^3$ of water is 15716.8 Joules.

Enthalpy change when 1 mole of solute is dissolved in 1.0000 $1000.00 cm^3$ of water is:

$\frac{15716.8 }{100.00 cm^3}\times 1000 cm^3=157168 J=157.168 kJ$

When 1 mole of the solute is dissolved in 1.0000 $dm^3$ of water, the change in enthalpy is 157.168 kJ/mol.

Alekssandra [3K] · Answer 2 · 2026-02-27T17:46:15+03:00

The result is: 16.56 kJ.
1) The mass of NH₄Cl is 5.35g.
For water, m(H₂O) = density(H₂O) · volume(H₂O) = 1g/cm³ · 100cm³ = 100g.
The temperature change, ΔT, is 25.55°C - 21.79°C = 3.76°C.
Calculating heat, Q = m(solution) · C(specific heat of water) · ΔT.
This gives Q = 105.35g · 4.18 J/g·°C · 3.76°C = 1655.76J.
2) Here, m(NH₄Cl) = 1mol · 53.5g/mol = 53.5g.
For water again, m(H₂O) = density(H₂O) · volume(H₂O) = 1g/cm³ · 1000cm³ = 1000g.
Thus, m(solution) totals 1053.5g, which is ten times the mass of the first solution.
Now, Q = 10 · 1655.76J = 16557.6J = 16.56 kJ.