An aqueous KNO3 solution using 72.5 g of KNO3 diluted to a total soultion volume

Question

An aqueous KNO3 solution using 72.5 g of KNO3 diluted to a total soultion volume of 2.00 L. Calculate the molarity, molality, and mass percent of the solution (assume density of 1.05 g/mL).

A sioxin-contaminated water source contains 0.085% dioxin by mass. How much dioxin is present in 2.5L of this water? Assume a density of 1.00 g/mL.

A solution is prepared by dissolving 20.2 mL of methanol (CH3OH)  in 100.0 mL of water at 25 degrees celsius. The final volue of the solution is 118 mL. The densitites of methanol and water at this temperature are 0.782 g/mL and 1.00 g/mL, respectively. For solution, caculate the concetation of each unit.

a. molarity

b. molality

c. percent by mass

d. mole fraction

e. mole percent

Explanation / Answer

moles KNO3 = 72.5 g /101.1 g/mol=0.717

M = 0.717/ 2 L=0.358

mass solution = 1.05 g/mL x 2000 mL = 2100 g
mass water = 2100 - 72.5=1412.5 g => 2.0275 Kg
m = 0.717 mol/ 2.0275 =0.3536

% by mass = 72.5 x 100 / 2100 =3.452

2)

2.5 L = 2500 mL

mass solution = 2500 mL x 1 g/mL = 2500 g

0.085 % = mass solute 100 / mass solution = mass solute x 100 / 2500

mass solute = 2500 x 0.085 / 100=2.13 g

3)

molarity = moles solute / litres of total solution

molality = moles solute / kg of solvent

mass percent solute = grams solute / total g solution x 100/1

mole fraction = moles solute / total moles in solution

mole percent = moles solute / total moles in solution x 100/1

So you need convert the solute and solvent into all the things need for the equations above,


moles = mass / molar mass
density = mass / volume



Methanol first

volume methanol = 20.2 ml = 0.0202 L

mass methanol = density x volume
= 0.782 g/ml x 20.2 ml
= 15.8 g

moles methanol = mass / molar mass
= 15.8 g / 32.042 g/mol
= 0.493 moles

Now water

volume H2O = 100.0 ml = 0.1000 L

mass water = density x volume
= 1.00 g/ml x 100.0 ml
= 100.0 g

moles H2O = mass / molar mass
= 100.0 g / 18.016 g/mol
= 5.55 moles

Total solution

volume = 118 ml = 0.118 L

mass = mass MeOH + mass CH3OH
= 15.8 g + 100.0 g
= 115.8 g

total moles = moles CH3OH + moles H2O = 0.493 mol + 5.55 moles
= 6.043 mol


Molarity = moles CH3OH / litres total solution
M = 0.493 mol / 0.118 L
= 4.18 M

Molality = moles solute / kg solvent
= moles CH3OH / kg H2O
= 0.493 mol / 0.1000 kg
= 4.93 m

percent mass CH3OH = mass CH3OH / total mass solution x 100/1
= 15.8 g / 115.8 g x 100/1
= 13.6 %

mole fraction CH3OH = moles CH3OH / total moles in solution
= 0.493mol / 6.043 mol
= 0.0816

mole percent CH3OH = moles CH3OH / total moles solution x100 /1
= 0.493 mol / 6.043 mol x 100 /1
= 8.16 %

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