A 3.0g sample of NaCl is added to a Styrofoam cup of water, and the change in water temperature is 6.5∘C . The heat of solution of NaCl is 3.76 kJ/mol.

What is the mass (in g) of water in the Styrofoam cup?

How much heat, q, is evolved ? That's

q = 3.76 kJ/mol x (3.0g NaCl/58.46 g NaCl) = ?. Then
q = mass H2O x specific heat H2O x delta T.
Substitute and solve for mass H2O

To solve this problem, we need to use the concept of heat of solution and heat capacity. The heat of solution (ΔHsoln) is the amount of heat released or absorbed when one mole of a solute is dissolved in a solvent.

First, we need to calculate the heat released when 3.0 g of NaCl is dissolved.

1. Convert the mass of NaCl to moles:
Number of moles = mass / molar mass

The molar mass of NaCl (sodium chloride) is 58.44 g/mol.

Number of moles of NaCl = 3.0 g / 58.44 g/mol = 0.051 mol

2. Calculate the heat released:
Heat released = ΔHsoln * number of moles of NaCl

The heat of solution (ΔHsoln) for NaCl is given as 3.76 kJ/mol. We need to convert it to joules:
ΔHsoln = 3.76 kJ/mol * 1000 J/ 1 kJ = 3760 J/mol

Heat released = 3760 J/mol * 0.051 mol = 192.76 J

Now, we need to find the heat capacity of the water in the Styrofoam cup. The heat capacity (C) is the amount of heat required to raise the temperature of a substance by 1 degree Celsius.

3. Calculate the heat capacity of the water in the cup:
Heat capacity = heat released / temperature change

The temperature change is given as 6.5 °C.

Heat capacity = 192.76 J / 6.5 °C = 29.65 J/°C

Lastly, we can determine the mass of the water in the Styrofoam cup by using the heat capacity and the specific heat capacity of water.

4. Use the formula:
Heat capacity = mass of water * specific heat capacity of water

The specific heat capacity of water is approximately 4.18 J/g°C.

Mass of water = heat capacity / specific heat capacity of water

Mass of water = 29.65 J/°C / 4.18 J/g°C = 7.11 g

Therefore, the mass of water in the Styrofoam cup is approximately 7.11 g.

To determine the mass of water in the Styrofoam cup, we need to use the information provided.

We know that a 3.0g sample of NaCl is added to the water, and the change in water temperature is 6.5∘C. Additionally, the heat of solution of NaCl is given as 3.76 kJ/mol.

To solve this problem, we can use the equation:

q = m * C * ΔT

where:
q = heat (in joules)
m = mass (in grams)
C = specific heat capacity (in J/g∘C)
ΔT = change in temperature (in ∘C)

First, we need to convert the heat of solution from kJ/mol to J/g. We can do this by multiplying the given value by 1,000 to convert kJ to J and then dividing by the molar mass of NaCl. The molar mass of NaCl is 58.44 g/mol.

Heat of solution in J/g = (3.76 kJ/mol * 1000 J/kJ) / 58.44 g/mol

Now, we can plug the values into the equation and solve for the mass of water:

q = m * C * ΔT

The heat (q) is equal to the heat of solution, which we just calculated.

m = (q) / (C * ΔT)

C represents the specific heat capacity of water, which is approximately 4.18 J/g∘C.

ΔT is the change in temperature, given as 6.5∘C.

Now we can substitute the values into the equation:

m = (3.76 kJ/mol * 1000 J/kJ) / (4.18 J/g∘C * 6.5∘C)

By canceling out units and performing the calculation, we can find the mass of water in the Styrofoam cup.