Question

A 7. 80 g sample of a nonvolatile nonelectrolyte blue crystalline solid is dissolved in 41. 0 g of acetic acid, producing a solution that bolis at 119. 7oC. Find the molar mass of the blue solid in g/mol

Answer 1

To find the molar mass of the blue solid in g/mol, we can use the formula: Molar mass (g/mol) = (mass of the sample in g) / (moles of the sample). First, we need to find the moles of the sample by using the formula: Moles of the sample = (mass of the sample in g) / (molar mass of the sample in g/mol). Given the mass of the sample and the boiling point elevation of the solution, we can calculate the molar mass of the sample.

Step-by-step explanation:

To find the molar mass of the blue solid in g/mol, we can use the formula:

Molar mass (g/mol) = (mass of the sample in g) / (moles of the sample)

First, we need to find the moles of the sample. We can do this by using the formula:

Moles of the sample = (mass of the sample in g) / (molar mass of the sample in g/mol)

Given that the sample mass is 7.80 g, we need to determine the molar mass of the sample. To do this, we can use the boiling point elevation of the solution. The boiling point elevation is the difference between the boiling point of the solution and the boiling point of the pure solvent.

Boiling point elevation (ΔTb) is given by the formula:

ΔTb = i * K_b * m

Where i is the Van't Hoff factor, K_b is the molal boiling point elevation constant, and m is the molality of the solution.

Since acetic acid is a nonvolatile nonelectrolyte, its Van't Hoff factor (i) is 1.

The molal boiling point elevation constant (K_b) for acetic acid is 3.07 °C/m.

Given that the solution boils at 119.7 °C and the boiling point of pure acetic acid is 118.1 °C, we can calculate the boiling point elevation:

ΔTb = 119.7 °C - 118.1 °C = 1.6 °C

Now we can find the molality of the solution:

m = (ΔTb) / (K_b) = 1.6 °C / 3.07 °C/m = 0.521 mol/kg

Finally, we can calculate the moles of the sample:

Moles of the sample = (mass of the sample in g) / (molar mass of the sample in g/mol) = 7.80 g / (0.521 mol/kg * 41.0 g/mol * 0.041 kg/g) = 9.86 mol

Now we can find the molar mass of the sample:

Molar mass (g/mol) = (mass of the sample in g) / (moles of the sample) = 7.80 g / 9.86 mol = 0.790 g/mol

Answer 2

To find the molar mass of the blue solid, we can use the formula molar mass = (mass of solute) / (moles of solute). The moles of the solute can be calculated using the mass of the sample and the molar mass of the solute. By using the equation for boiling point elevation, we can determine the molality of the solution. Substituting the values into the equation, we can calculate the molar mass of the solute.

Step-by-step explanation:

To find the molar mass of the blue solid in the solution, we can use the formula:

molar mass = (mass of solute) / (moles of solute)

First, we need to find the moles of the solute. We can use the formula:

moles of solute = (mass of solute) / (molar mass of solute)

Given that the mass of the sample is 7.80 g and it is dissolved in 41.0 g of acetic acid, we can calculate the moles of the solute:

moles of solute = (7.80 g) / (molar mass of solute)

Next, we need to calculate the molar mass of the solute. To do this, we need to use the equation for boiling point elevation:

ΔT = K * m * i

Where ΔT is the boiling point elevation, K is the boiling point elevation constant, m is the molality of the solution, and i is the van't Hoff factor.

In this case, we know that the solution boils at 119.7oC, so ΔT = 119.7oC - 100.0oC = 19.7oC. We also know that acetic acid is a nonelectrolyte, so i = 1.

We can rearrange the equation to solve for m:

m = ΔT / (K * i)

Now, we can substitute the values into the equation:

m = 19.7oC / (0.52oC/m * 1)

Calculating the value of m, we find:

m = 37.9 mol/kg

Finally, we can substitute the value of m into the moles of solute equation to solve for the molar mass of the solute:

37.9 mol = (7.80 g) / (molar mass of solute)

Solving for the molar mass, we find:

molar mass of solute = (7.80 g) / (37.9 mol) = 0.206 g/mol