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
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.
are these rite
molality 1.125
molarity-1.2
mole fraction-.34
mole percent- 34%
Check your 2-7-11,10:46am post.
Check your 2-7-11,10:51 post
To determine the molality, molarity, mole percent, % mass, and ppm by mass of the solution, as well as the freezing point and boiling point, we need to go through several steps. I'll explain each calculation in detail.
Step 1: Calculate the number of moles of solute (KCl).
Given:
Mass of KCl = 115 grams
To calculate the moles, we need the molar mass of KCl.
Molar mass of KCl = 39.10 g/mol (atomic mass of K) + 35.45 g/mol (atomic mass of Cl)
= 74.55 g/mol
Number of moles of KCl = Mass of KCl / Molar mass of KCl
= 115 g / 74.55 g/mol
≈ 1.54 moles (rounded to two decimal places)
Step 2: Calculate the molality (m).
Given:
Mass of solvent (water) = 750 ml (milliliters) = 0.75 L (liters)
Molality (m) = Moles of solute / Mass of solvent (in kg)
= 1.54 moles / 0.75 kg
≈ 2.05 mol/kg (rounded to two decimal places)
Step 3: Calculate the molarity (M).
Given:
Volume of the solution (water + KCl) = 750 ml = 0.75 L
Molarity (M) = Moles of solute / Volume of solution (in L)
= 1.54 moles / 0.75 L
= 2.05 M
Step 4: Calculate the mole percent.
Mole percent of KCl = (Moles of KCl / Total moles in the solution) * 100
= (1.54 moles / (1.54 moles + 55.55 moles)) * 100
≈ 2.71% (rounded to two decimal places)
Step 5: Calculate the % mass.
% Mass of KCl = (Mass of KCl / Mass of solution) * 100
= (115 g / (115 g + 750 g)) * 100
≈ 13.28% (rounded to two decimal places)
Step 6: Calculate ppm by mass (parts per million by mass).
ppm by mass = (% Mass of KCl / 10,000)
= (13.28% / 10,000)
≈ 1328 ppm (rounded to the nearest whole number)
Step 7: Calculate the freezing point depression (ΔTf).
Given:
Kf (freezing point constant) = 1.86 °C/m
ΔTf = Kf * m
= 1.86 °C/m * 2.05 mol/kg
≈ 3.80 °C (rounded to two decimal places)
The freezing point of pure water is 0 °C, so the freezing point of the solution would be 0 °C - 3.80 °C = -3.80 °C
Step 8: Calculate the boiling point elevation (ΔTb).
Given:
Kb (boiling point constant) = 0.0512 °C/m
ΔTb = Kb * m
= 0.0512 °C/m * 2.05 mol/kg
≈ 0.105 °C (rounded to three decimal places)
The boiling point of pure water is 100 °C, so the boiling point of the solution would be 100 °C + 0.105 °C = 100.105 °C.
So, the freezing point of the solution would be approximately -3.80 °C, and the boiling point would be approximately 100.105 °C.