Question

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

Answer 1

Answer is: 16,56 kJ.
1) m(NH₄Cl) = 5,35g.
m(H₂O) = d(H₂O) · V(H₂O) = 1g/cm³ · 100cm³ = 100g.
ΔT = 25,55°C - 21,79°C = 3,76°C.
Q = m(solution) · C(specific heat capacity of water) ·ΔT.
Q = 105,35g · 4,18 J/g·°C · 3,76°C = 1655,76J.
2) m(NH₄Cl) = 1mol · 53,5g/mol = 53,5g.
m(water) = d(H₂O) · V(H₂O) = 1g/cm³ · 1000cm³ = 1000g.
m(solution) = 1053,5g, ten times more than first solutn.
Q = 10 · 1655,76J = 16557,6J  = 16,56 kJ.

Answer 2

Answer:

The enthalpy change that would occur when 1 mole of the solute is added to 1.0000 $dm^3$ of water is 157.168 kJ/mol.

Explanation:

Amount of ammonium chloride added = 5.350 g

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

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

Q = -Q'

Volume of water,V = $100.00 cm^3=100.00 mL$

Mass of water = m

Density of water ,d= 1 g/mL

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

Change in temperature of the water  =ΔT = 21.79°C- 25.55°C = -3.76°C

Specific heat capacity of water = c = 4.18J/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

0.1 mole of solute absorbed 1571.68 Joules of heat from $100.00 cm^3$ of water .

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

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

$1 dm^3=1000 cm^3$

15716.8 Joule of heat is absorbed when 1 mole of solute is dissolved in $100.00 cm^3$ of water

Heat absorbed when 1 mole of solute is dissolved in $1000.00 cm^3$ of water :

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

The enthalpy change that would occur when 1 mole of the solute is added to 1.0000 $dm^3$ of water is 157.168 kJ/mol.