115 grams of KCl is dissolved in 750 ml of water. what are the molality, molarity, mole percent, % mass, ppm by mass. what would the freezing point and boiling point of that solution assuming Kf of water is 1.86 degree celsius/m and Kb is .0512 degree celsius/m

chemisty - DrBob222, Monday, February 7, 2011 at 11:30pm
I'll start you off with the first one.
molality = moles/kg solvent.
moles = grams/molar mass
kg solvent = 0.750 kg
molar mass KCl about 74.5
Therefore, moles KCl = 115/74.5 = x moles
molality = x moles KCl/0.750 kg. = zz molality.
I'll be glad to help you do the remainder of the problem but I won't work each one for you.

chemisty - Anonymous, Tuesday, February 8, 2011 at 9:34am
are these rite

molality 1.125

molarity-1.2

mole fraction-.34

mole percent- 34%

Let's go through the calculations step by step:

1. Molality:
Molality is defined as the number of moles of solute per kilogram of solvent.

Moles of KCl = mass / molar mass
= 115 g / 74.5 g/mol
= 1.543 mol

molality = moles KCl / kg solvent
= 1.543 mol / 0.750 kg
= 2.057 mol/kg

Therefore, the molality is 2.057 mol/kg

2. Molarity:
Molarity is defined as the number of moles of solute per liter of solution.

Volume of solution = 750 ml = 0.750 L

Moles of KCl = mass / molar mass
= 115 g / 74.5 g/mol
= 1.543 mol

Molarity = moles KCl / liters of solution
= 1.543 mol / 0.750 L
= 2.058 M

Therefore, the molarity is 2.058 M

3. Mole Percent:
Mole percent is the percentage of the total moles accounted for by a given component.

Total moles = moles KCl

Mole percent of KCl = (moles KCl / total moles) * 100
= (1.543 mol / 1.543 mol) * 100
= 100%

Therefore, the mole percent of KCl is 100%

4. % Mass:
% Mass is the mass of the solute per 100 g of solution.

% Mass = (mass of KCl / mass of solution) * 100
= (115 g / (115 g + 750 g)) * 100
= 13.27%

Therefore, the % mass is 13.27%

5. ppm by mass:
ppm (parts per million) is a way of expressing the concentration of a solute in a solution.

ppm = (mass of solute / mass of solution) * 10^6
= (115 g / (115 g + 750 g)) * 10^6
= 126,844 ppm

Therefore, the ppm by mass is 126,844 ppm

Now, let's move on to the freezing point and boiling point calculations:

Freezing point depression:
∆Tf = Kf * molality
= 1.86°C/m * 2.057 mol/kg
= 3.8182°C

Freezing point of solution = freezing point of pure water - ∆Tf
= 0°C - 3.8182°C
= -3.8182°C

Therefore, the freezing point of the solution is -3.8182°C.

Boiling point elevation:
∆Tb = Kb * molality
= 0.0512°C/m * 2.057 mol/kg
= 0.1052°C

Boiling point of solution = boiling point of pure water + ∆Tb
= 100°C + 0.1052°C
= 100.1052°C

Therefore, the boiling point of the solution is 100.1052°C.

To find the molality, molarity, mole fraction, and mole percent of a solution, you need to know the mass and volume of the solute and solvent and the molar mass of the solute.

1. Molality (m):
Molality is defined as moles of solute per kilogram of solvent.
To find molality, you need to first calculate the number of moles of solute (KCl) using the formula:
moles = grams / molar mass
In this case, the mass of KCl is 115 grams and the molar mass of KCl is approximately 74.5 g/mol.
Moles of KCl = 115 g / 74.5 g/mol = 1.547 moles
Next, divide the moles of KCl by the mass of the solvent (water) in kilograms:
Molality = moles / kg of solvent
Given that the volume of water is 750 mL, which is equivalent to 0.750 L or 0.750 kg:
Molality = 1.547 moles / 0.750 kg = 2.063 mol/kg

2. Molarity (M):
Molarity is defined as moles of solute per liter of solution.
To find molarity, you need to divide the moles of solute by the volume of the solution in liters.
Molarity = moles / liters of solution
Given that the volume of the solution is 0.750 L:
Molarity = 1.547 moles / 0.750 L = 2.063 M

3. Mole fraction (χ):
Mole fraction is the ratio of moles of a component to the total moles of all components in the solution.
To find mole fraction, you need to calculate the moles of each component (solute and solvent) and then divide the moles of the solute by the total moles.
In this case, the moles of KCl is 1.547, and the moles of water can be calculated using the formula:
mass of water = volume of water * density of water
Here, the density of water is approximately 1 g/mL.
mass of water = 0.750 kg - 0.115 g (mass of KCl)
mass of water = 0.635 kg
Moles of water = 0.635 kg / 18 g/mol (molar mass of water) = 0.0353 moles
Total moles = 1.547 moles (KCl) + 0.0353 moles (water) = 1.5823 moles
Mole fraction of KCl = 1.547 moles / 1.5823 moles = 0.977

4. Mole percent:
Mole percent is the mole fraction multiplied by 100.
Mole percent = mole fraction * 100
Mole percent of KCl = 0.977 * 100 = 97.7%

Please note that the calculations for molarity, mole fraction, and mole percent were not provided in the question, so it is important to double-check the calculations to ensure accuracy.

Regarding the freezing point and boiling point of the solution:
The freezing point depression (change in freezing point) can be calculated using the formula:
ΔTf = Kf * molality
where ΔTf is the change in freezing point, and Kf is the freezing point depression constant for water (given as 1.86°C/m).
In this case, molality = 2.063 mol/kg.
ΔTf = 1.86°C/m * 2.063 mol/kg = 3.83°C (approx.)

The boiling point elevation (change in boiling point) can be calculated using the formula:
ΔTb = Kb * molality
where ΔTb is the change in boiling point, and Kb is the boiling point elevation constant for water (given as 0.0512°C/m).
ΔTb = 0.0512°C/m * 2.063 mol/kg = 0.105°C (approx.)

Therefore, the freezing point of the solution would be 0°C - 3.83°C = -3.83°C (approximately), and the boiling point of the solution would be 100°C + 0.105°C = 100.105°C (approximately).

molality. no. I worked the problem for youl. Did you punch in the wrong numbers?

115/74.5 = 1.54 and 1.54/0.750 = aboput 2.06m
Molarity can't be done without the density of the solution.

mole fraction = 0.36
mole % = 36%