the density of an aqueous solution containing 10.0% by mass ethanol is 0.984 g/ml. calculate molality and molarity. what volume of the solution would contain 1.15 g ethanol ?

Molality(m) = moles solute/Kilograms Solvent

For a 10.0% w/w aqueous solution of EtOH(f.wt. = 46g/mol) => 10gms EtOH/100gms Soln. Using this ratio as solute-in-solution base solution, then the data needed for molality would be 10gms EtOH in (100-90)gms solvent => molality = ((10/46)mole EtOH)/0.09Kg Solvent = 2.42 molal in EtOH...

Molarity(M) = Moles Solute/Liter Solution

For the same 10% w/w EtOH solution => 10gm EtOH/100gm Solution:
=> [EtOH]=[(10/46)mole EtOH]/[(100gm soln/0.984g/ml)/(1000ml/L)]
=(0.2174 mole EtOH/0.1016L soln)
=2.10 Molar in EtOH

You probably already know this, but molality(m) is used in solution type problems where the solution goes through a temperature change history, whereas, Molarity(M) based solutions are bound to a constant temperature. Since Molarity = moles solute/Liters of Solution, heating or cooling a 'Molar Solution' changes the solute concentration b/c volume expands or contracts changing the concentration. The value of Molality(m) = (moles solute/Kg Solvent) is not the same if the temperature of the solution changes.Boiling point elevation and Freezing Pt depression type problems are molality type cases in point. Just a polite FYI if you needed it. Enjoy.

Correction... Molality(m) remains the same if temp is changed... (next to last sentence). remove the 'not'. If there's an edit feature on this site, I don't know how to use it. sorry bout that.

To calculate the molality (m) and molarity (M) of the solution, we need to know the molar mass of ethanol (C2H5OH), which is 46.068 g/mol.

First, let's calculate the molality (m):
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

Given:
Mass of ethanol = 10.0% = 10.0 g
Density of the solution = 0.984 g/mL

Since the density is given, we can calculate the volume of the solution using the formula:
Volume of solution (V) = mass of solution / density

Mass of solution = Mass of ethanol / (percentage of ethanol / 100)
Mass of solution = 10.0 g / (10.0 / 100) = 100 g

V = 100 g / 0.984 g/mL ≈ 101.63 mL

Now let's convert the volume of the solution to kilograms:
1 mL = 1 cm³ = 1 gram
1000 mL = 1000 grams = 1 kilogram

Volume in kilograms (V') = 101.63 mL * 1 g/mL * (1 kg / 1000 g) = 0.10163 kg

Now we can calculate the molality:
Molality (m) = Moles of solute / Kilograms of solvent

Moles of solute = Mass of solute / Molar mass of ethanol
Moles of solute = 10.0 g / 46.068 g/mol ≈ 0.2173 mol

Molality (m) = 0.2173 mol / 0.10163 kg ≈ 2.135 mol/kg

Next, let's calculate the molarity (M):
Molarity (M) is defined as the number of moles of solute per liter of solution.

Moles of solute = Mass of solute / Molar mass of ethanol
Moles of solute = 1.15 g / 46.068 g/mol ≈ 0.0250 mol

Volume in liters (V'') = Volume in milliliters / 1000 = 1.15 mL / 1000 mL/L = 0.00115 L

Molarity (M) = Moles of solute / Volume of solution in liters
Molarity (M) = 0.0250 mol / 0.00115 L ≈ 21.7 M

Finally, to calculate the volume of the solution that would contain 1.15 g of ethanol, we can rearrange the equation for molarity as follows:
Moles of solute = Molarity (M) * Volume of solution in liters
0.0250 mol = 21.7 M * V
V = 0.0250 mol / 21.7 M ≈ 0.00115 L

Therefore, 1.15 g of ethanol would be contained in approximately 0.00115 liters (or 1.15 mL) of the solution.

To calculate the molality and molarity of the solution, we need to use the given information about the mass percent and density.

First, let's calculate the molality:

Molality (m) is defined as the number of moles of solute per kilogram of solvent.

1. Calculate the mass of the solvent:
- Given that the solution contains 10.0% by mass ethanol, the remaining 90.0% is water (the solvent).
- Assume we have 100 grams of the solution, which means there is 10 grams of ethanol and 90 grams of water.

2. Convert the mass of water into kilograms:
- 90 grams of water is equal to 90/1000 = 0.09 kilograms.

3. Calculate the number of moles of ethanol:
- The molar mass of ethanol (C2H5OH) is 46.07 g/mol.
- Convert the mass of ethanol into moles:
- 10 grams of ethanol is equal to 10/46.07 = 0.217 moles.

4. Calculate the molality:
- Molality (m) = moles of ethanol / mass of solvent in kg
- Molality (m) = 0.217 moles / 0.09 kg = 2.41 mol/kg.

Next, let's calculate the molarity:

Molarity (M) is defined as the number of moles of solute per liter of solution.

1. Calculate the volume of the solution:
- Given that the solution has a density of 0.984 g/ml, this means it has a mass of 0.984 grams per milliliter.
- To find the volume containing 1.15 grams of ethanol, divide the given mass by the density:
- Volume = mass / density = 1.15 g / 0.984 g/ml ≈ 1.17 ml.

2. Convert the volume to liters:
- 1.17 ml is equal to 1.17/1000 = 0.00117 liters.

3. Calculate the number of moles of ethanol:
- Convert the given mass of ethanol into moles:
- 1.15 grams of ethanol is equal to 1.15/46.07 = 0.025 moles.

4. Calculate the molarity:
- Molarity (M) = moles of ethanol / volume of solution in liters
- Molarity (M) = 0.025 moles / 0.00117 L ≈ 21.37 M.

Therefore, the molality of the solution is approximately 2.41 mol/kg, and the molarity is approximately 21.37 M.