When 4.76 g of a solid mixture composed of NH4Cl and CaCl2 was dissolved in 100.0 mL of water, the temperature of the water rose by 4.73°C. How many grams of each substance was in the mixture?
I looked up the enthalpy of solvation for NH4Cl and CaCl2. I can work this problem for you now. If you still need help, please repost as a new question and I will get to it.
Do you have any other data given such as heat solvation etc?
4 grams
What is the answer ? And how
To find out how many grams of each substance were in the mixture, we can use the concepts of heat capacity and heat transfer. The change in temperature of the water is directly related to the amount of heat transferred to it.
First, we need to calculate the heat transfer to the water. The equation to calculate heat transfer is:
Q = mcΔT
Where Q is the heat transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
In this case, we know the change in temperature (ΔT) is 4.73°C, the mass of water (m) is 100.0 mL, and the specific heat capacity of water (c) is 4.18 J/(g°C).
Converting the mass of water from mL to grams, we use the density of water, which is 1.00 g/mL. Therefore, 100.0 mL of water is equivalent to 100.0 g.
Plugging the values into the equation, we get:
Q = (100.0 g)(4.18 J/(g°C))(4.73°C)
Q = 1978.14 J
Next, we need to determine the heat transfer by the dissolution of the solid mixture. The heat transfer by dissolution is equal to the heat absorbed by the water.
Since we have the mass of the solid mixture (4.76 g), but we don't know the exact quantities of NH4Cl and CaCl2, we can assign variables to represent their masses.
Let x be the mass of NH4Cl and y be the mass of CaCl2.
Since the heat transfer is proportional to the mass of the substance, we can write:
Q = qx + py
Where q is the heat capacity of NH4Cl and p is the heat capacity of CaCl2.
We need to know the specific heat capacity of NH4Cl and CaCl2 to find the values of q and p, respectively. Assuming the specific heat capacities of NH4Cl and CaCl2 are 4.18 J/(g°C) (same as water), we can calculate q and p.
Q = qx + py
1978.14 J = (4.18 J/(g°C))x + (4.18 J/(g°C))y
Simplifying the equation, we have:
1978.14 = 4.18x + 4.18y
Now we have an equation with two variables. To solve for x and y, we need another equation. This new equation is based on the mass of the solid mixture.
The mass of the solid mixture is equal to the sum of the masses of NH4Cl and CaCl2:
4.76 g = x + y
We now have two equations:
1978.14 = 4.18x + 4.18y
4.76 = x + y
Using these two equations, we can solve for x and y using simultaneous equations or substitution method.
Substituting the value of y from the second equation into the first equation, we get:
1978.14 = 4.18x + 4.18(4.76 - x)
1978.14 = 4.18x + 19.9248 - 4.18x
1978.14 - 19.9248 = 4.18(x - x)
1958.2152 = 0
This equation is not valid, which means there is no solution that satisfies both equations.