What mass (in grams) of CaCl2(s) must be dissolved in pure water at 10.0 deg Celsius to make a 26.4 mL solution and to increase the solution temperature to 16.4 deg Celsius? Assume that there is no heat loss from the solution and that the solution has the same physical properties as pure water.
Data:
DHsol(CaCl2)(s)) = - 82.8 kJ.mol-1
specific heat of water = 4.184 J.g-1.deg-1
density of water = 1.00 g.cm-3
molar mass of CaCl2 = 111 g/mol
To solve this problem, we need to use the equation:
q = m × s × ΔT
where:
q is the heat absorbed or released by the solution (in joules),
m is the mass of the solution (in grams),
s is the specific heat of water (in J.g-1.deg-1),
and ΔT is the change in temperature (in degrees Celsius).
First, let's calculate the heat absorbed by the solution using the equation:
q = m × s × ΔT
q = (26.4 mL) × (1.00 g/cm^3) × (4.184 J.g-1.deg-1) × (16.4 - 10.0) deg Celsius
Note: We convert the volume to mass using the density of water (1.00 g/cm^3), and the specific heat of water (4.184 J.g-1.deg-1).
q = (26.4 g) × (4.184 J.g-1.deg-1) × (6.4 deg Celsius)
q = 689.1648 J
Next, let's calculate the moles of CaCl2 needed to release the same amount of heat:
DHsol(CaCl2)(s) = - 82.8 kJ.mol-1 = -82,800 J.mol-1
moles of CaCl2 = q / DHsol(CaCl2)(s)
moles of CaCl2 = 689.1648 J / -82,800 J.mol-1
moles of CaCl2 = -0.00832 mol
Lastly, let's convert moles of CaCl2 to mass using the molar mass of CaCl2:
mass of CaCl2 = moles of CaCl2 × molar mass of CaCl2
mass of CaCl2 = -0.00832 mol × 111 g/mol
mass of CaCl2 = -0.92272 g
Since mass cannot be negative, we can conclude that no dissolving of CaCl2 is required to achieve the desired temperature increase.
To solve this problem, we need to calculate the amount of heat absorbed by the water to increase the temperature from 10.0°C to 16.4°C. This heat will come from the dissolution of CaCl2 in the water.
First, let's calculate the heat required to increase the temperature of the water:
Q = m * C * ΔT
where Q is the heat required, m is the mass of water, C is the specific heat of water, and ΔT is the change in temperature.
Given:
m = 26.4 mL = 26.4 g (since the density of water is 1.00 g/cm³)
C = 4.184 J/g°C
ΔT = 16.4°C - 10.0°C = 6.4°C
Plugging in the values:
Q = 26.4 g * 4.184 J/g°C * 6.4°C
Q = 699.4176 J
Now, let's calculate the moles of CaCl2 needed to release this amount of heat:
DHsol(CaCl2) = -82.8 kJ/mol = -82.8 * 1000 J/mol
molar mass of CaCl2 = 111 g/mol
ΔH = n * DHsol(CaCl2)
Solving for n (moles of CaCl2):
n = ΔH / DHsol(CaCl2)
n = 699.4176 J / -82.8 * 1000 J/mol
n = -8.44 * 10^-3 mol
Since 1 mole of CaCl2 has a molar mass of 111 g/mol, the mass of CaCl2 required is:
mass = n * molar mass of CaCl2
mass = -8.44 * 10^-3 mol * 111 g/mol
mass = -0.936 g
However, mass cannot be negative, so we took the absolute value:
mass = 0.936 g
Therefore, 0.936 grams of CaCl2 must be dissolved in pure water at 10.0°C to make a 26.4 mL solution and increase the temperature to 16.4°C.