Download Calculating Concentration and Volume of Solutions and more Lecture notes Chemistry in PDF only on Docsity!
Reactions in Aqueous Solution
Chapter 4
Solutions and concentration
- A solution is a homogeneous mixture of 2 or more substances (gas, liquid, or solid) in a single phase and it contains a solute (the substance thatcontains a solute (the substance that is dissolved in a solvent) and a solvent (a liquid in which a substance is dissolved).
- When the solvent is water, the solution is said to be aqueous (aq).
Concentration of solution can be expressed in different ways:
2
Concentration of solution can be expressed in different ways:
volumeofsolution (liter) Molarity( M) molesofsolute
Weight% weightweightofofsolutionsolute x 100
**Calculate the mass required to prepare a 250 mL 0.01 M solution of KMnO 4? Convert 250 ml to L (250/1000 = 0.250 L) Using the formula:
moles = molarity x volume
= 0.01 mol/L x 0.250 L =0.0025 mol Mass = # moles x molar mass Molar mass of KMnO4 = 158.0 g/mole Mass of KMnO 4 needed = 0.0025 mol x 158.0 g/mole = 0.395 g of KMnO 4 So, weigh 0.395 g of KMnO 4 and dissolve them in 250 ml volumetric flask.**
3
If a solution contains 0.035 moles solute in 2.0 L of water, what is the molarity? Molarity (M) = moles of solute / volume of solution (liter) = 0.035 moles / 2.0 L = 1.8 x 10 -2^ M
Dilution of Concentrated Solutions
- When we dilute a solution by mixing it with more solvent, the amount of solute present does not change, but the total volume and the concentration of the solution do change.
- To calculate the molarity after dilution, we can use the following formula:
(Molarity x Volume) (^) before dilution = (Molarity x Volume)after dilutio Mi x Vi = Mf x Vf
4
How many milliliters of 18.0 M H 2 SO 4 are required to prepare 1.00 L of a 0.900 M solution of H 2 SO 4? Using the formula: M (^) i x V (^) i = M (^) f x V (^) f M (^) i = 18.0 M, V (^) i=?? And M (^) f = 0.900 M, V (^) f = 1.00 L So,
- 0500 L 50. 0 mL 18.0M
0.900Mx1.00L M
M xV V 1
2 2 1
According to the reaction: Ba(OH) (^) 2 (aq) + 2 HNO (^) 3 (aq) → Ba(NO 3 ) (^) 2 (aq) + 2 H 2 O (^) (l) What volume of 0.5M HNO 3 is required to react with 41.77 mL of 0.1603 M Ba(OH) 2?
From the chemical equation: Ba(OH)Ba(OH) 2 2 (aq) (aq) + 2 HNO2 HNO (^) 3 (aq) (^) 3 (aq) → Ba(NOBa(NO 3 3 ) ) 2 2 (aq) (aq) + 2 H2 H 2 2 O O (^) (l)(l)
2 Moles of HNO 3 react with one mole of Ba(OH) (^2)
moles of Ba(OH) 2 = molarity X volume of solution
= 0.1603 M X (41.77/1000) L = 6.696 x 10 -3^ mol
The moles of HNO 3 which reacted = 2 x 6.696 x 10 -3^ = 13.39 x 10 -3^ mol
l f HNO l it X l f l ti
7
moles of HNO 3 = molarity X volume of solution
13.39 x 10 -3^ mol = 0.5 M X V V= 0.02678 L = 26.78 mL
Mole Fraction
- The mole fraction, X, is the concentration expression most like
percent composition yet includes mole ratio information.
- The mole fraction is the number of moles of one component of a
i t di id d b th t t l b f l i th i t
mole fraction (Xi ) =
ni nT
XA =
n (^) A
n A +n B
Example: A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C 2 H 6 , and 0.116 moles of C 3 H 8. What is the mole fraction of propane (C H )?
mixture divided by the total number of moles in the mixture.
8
What is the mole fraction of propane (C 3 H 8 )?
Xpropane =
