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CHEM-REVIEWER 2nd sem prelims

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CHEM REVIEWER
☺ MOLARITY
unit: M or mol/L
οƒ˜ Molar solution of a substance is defined as a solution
contains 1g molecular weight (1 mole of the solute in
1L solution) of the substance/liter of the solution
οƒ˜ 𝑀=
49𝑔
= 1.0π‘šπ‘œπ‘™/π‘˜π‘”
(98 𝑔/π‘šπ‘œπ‘™)(0.5π‘˜π‘”)
π‘š=
π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
𝐿 π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
π‘€π‘œπ‘™π‘Žπ‘Ÿπ‘–π‘‘π‘¦ =
π‘”π‘Ÿπ‘Žπ‘šπ‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
since π‘šπ‘œπ‘™π‘’π‘  = π‘šπ‘œπ‘™π‘’π‘π‘’π‘™π‘Žπ‘Ÿ π‘€π‘’π‘–π‘”β„Žπ‘‘
π‘”π‘Ÿπ‘Žπ‘šπ‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
(π‘€π‘Š π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’)(𝐿 π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’)
Example:
1. Calculate the molarity of 2.0 L of solution containing
5.0 moles of NaOH.
𝑀=
5.0π‘šπ‘œπ‘™ π‘ π‘œπ‘™π‘’π‘‘π‘’
= 2.5𝑀 π‘œπ‘Ÿ 2.5π‘šπ‘œπ‘™/𝐿
2.0𝐿 π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
☺ SOLUTIONS
οƒ˜ Homogenous mixtures of 2 or more substances
οƒ˜ Parts:
1) Solute – substance that dissolves in a solution
2) Solvent – dissolving medium
☺ MOLE FRACTION
οƒ˜ The number of moles of 1 component to the total
number of moles of all components present
π‘€π‘œπ‘™π‘’ πΉπ‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘†π‘œπ‘™π‘’π‘‘π‘’ =
π‘šπ‘œπ‘™π‘’ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’ + π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘
π‘€π‘œπ‘™π‘’ πΉπ‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘†π‘œπ‘™π‘£π‘’π‘›π‘‘ =
π‘šπ‘œπ‘™π‘’ π‘œπ‘“ π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘
π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’ + π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘
2. Calculate the volume of a 0.750M sulfuric acid
(H2SO4) solution containing 0.120 moles of solute.
0.120π‘šπ‘œπ‘™
0.750𝑀 =
π‘₯𝐿
π‘₯ = 0.160𝐿
Activity:
Example:
1. Calculate the molarity of a solution containing 10g of
sulfuric acid (MW = 98) in 500mL of (pakidugtungan
kulang sakin hehe)
Given:
οƒ˜ 10g sulfuric acid
10𝑔
𝑀 = (98)(0.5𝐿) = 0.204π‘šπ‘œπ‘™/𝐿
οƒ˜ 98 MW
οƒ˜ 500mL
π‘šπ‘œπ‘™π‘’π‘  𝐻2𝑂 =
2. Calculate the weight in grams of sulfuric acid in 2L of
0.100M solution
Given:
οƒ˜ 0.100M solution
π‘”π‘Ÿπ‘Žπ‘šπ‘  = (𝑀)(π‘€π‘Š)(𝐿)
οƒ˜ 2L
π‘”π‘Ÿπ‘Žπ‘šπ‘  = (0.100)(98)(2𝐿)
οƒ˜ 98 MW
π‘”π‘Ÿπ‘Žπ‘šπ‘  = 19.6𝑔
☺ MOLALITY
π‘š=
1. What are the mole fractions of solute and solvent in a
solution prepared by dissolving 98g sulfuric acid in
162 water?
98𝑔
π‘šπ‘œπ‘™π‘’π‘  𝐻2𝑆𝑂4 =
= 1π‘šπ‘œπ‘™ (π‘ π‘œπ‘™π‘’π‘‘π‘’)
98𝑔/π‘šπ‘œπ‘™
unit: m or mpl/kg
π‘šπ‘œπ‘™ π‘ π‘œπ‘™π‘’π‘‘π‘’
(π‘€π‘Š)(π‘˜π‘” π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘)
Example:
1. What is the molality of a solution in which 49g sulfuric
acid is dissolved in 500g H20?
162𝑔
= 9 π‘šπ‘œπ‘™ (π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘)
18𝑔/π‘šπ‘œπ‘™
π‘šπ‘œπ‘™π‘’ π‘“π‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› 𝐻2𝑆𝑂4 =
1
= 0.1
1+9
π‘šπ‘œπ‘™π‘’ π‘“π‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› 𝐻2𝑆𝑂4 =
9
= 0.9
1+9
☺ NORMALITY FRACTION
οƒ˜ Gram equivalent weight of solute per liter solution
Gram equivalent =
𝑁=
MW
π‘£π‘Žπ‘™π‘’π‘›π‘π‘’
π‘”π‘Ÿπ‘Žπ‘šπ‘  π‘ π‘œπ‘™π‘’π‘‘π‘’
(πΈπ‘Š)(πΏπ‘–π‘‘π‘’π‘Ÿπ‘  π‘†π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›)
Example:
2) Volume in volume (v/v) percent solution
1. Calculate the normality of a solution containing 2.45g
sulfuric acid in 2L solution
𝑁=
2.45𝑔
= 0.025 𝑁 π‘œπ‘Ÿ π‘’π‘žπ‘’π‘–π‘£/𝐿
98
βˆ—
2𝐿
2
☺ PERCENT SOLUTION
οƒ˜ Amount of solute in a solution can be measured as a
percentage of total value of the solute. The term
percent refers to parts of solute per 100 parts of the
solvent
1) Weight in volume (w/v) percent solution
οƒ˜ Volume = 100
𝑀
π‘”π‘Ÿπ‘Žπ‘šπ‘  π‘ π‘œπ‘™π‘’π‘‘π‘’
% =
π‘₯ 100
𝑣 π‘šπ‘™ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
οƒ˜ Or grams of solute per 100ml of solution
Example:
a) What is the w/v% of solution that has 25.0g of
NaCl dissolved into a total volume of %100ml?
%
𝑀
25.0𝑔 π‘π‘ŽπΆπ‘™
=
π‘₯ 100 = 25%
𝑣 100π‘šπ‘™ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
οƒ˜ Therefore, a solution with 25.0g of NaCl dissolved into
100ml of H20 has a %w/v concentration of 25%
b) Prepare 200ml of 10% NaOH
(10)(200)
100
= 20𝑔 π‘π‘Žπ‘‚π» π‘žπ‘’π‘Žπ‘›π‘‘π‘–π‘‘π‘¦ 𝑠𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑑 π‘‘π‘œ 200π‘šπ‘™
Activity:
a) How many grams of NaOH would be needed to
make a 40% w/v solution using deionized water as
the solvent?
(40𝑔)(100)
= 40𝑔
100
b) How many grams of CaCl2 would be needed to
make 10% w/v solution using deionized water as
the solvent?
(10)(100)
= 10𝑔
100
%
𝑣 π‘šπ‘™ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
=
π‘₯ 100
𝑣
100π‘šπ‘™
οƒ˜ Or ml of solute per 100ml of solution
οƒ˜ % v/v is similar to % w/w in that the total volume of
the solution is 100ml
οƒ˜ π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ π‘œπ‘“ π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘ = 100π‘šπ‘™ π‘‘π‘œπ‘‘π‘Žπ‘™ π‘£π‘œπ‘™π‘’π‘šπ‘’ βˆ’
π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
Example:
a) Prepare 200ml of 5% ammonium hydroxide
solution
𝑣
π‘šπ‘™ π‘ π‘œπ‘™π‘’π‘‘π‘’
% =
π‘₯ 100
𝑣 100π‘šπ‘™ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
π‘šπ‘™ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’ =
π‘₯=
%π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘› π‘‘π‘’π‘ π‘–π‘Ÿπ‘’π‘‘ π‘₯ π‘‘π‘œπ‘‘π‘Žπ‘™ π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘‘π‘’π‘ π‘–π‘Ÿπ‘’π‘‘
100
(5)(200)
= 10π‘šπ‘™
100
οƒ˜ 10ml ammonium hydroxide quantity sufficient to
200ml
b) How many ml of ethanol are needed to make a
75% v/v solution using 100ml deionized water as
solute?
(75)(100)
= 75
100
c) How many ml of water are necessary for the
solution?
π‘₯ = 100 βˆ’ 75
π‘₯ = 25
d) How many ml of HCl are needed to make a 10%
v/v solution using deionized water as the solvent?
(10)(100)
= 10
10
3) Weight in weight percent solution (w/w)
%
𝑀 π‘”π‘Ÿπ‘Žπ‘šπ‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
=
π‘₯ 100
𝑀 100𝑔 π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
Example:
a) Prepare 100ml of 10% ammonium hydroxide solution
from the concentrated reagents in which 1ml of NaOH
contains 3.97g pure ammonia.
(3.97)(10)(100)
π‘₯=
= 39.7π‘šπ‘™
100
οƒ˜ 39.7ml concentrated ammonium hydroxide quantity
sufficient to 100ml
Activity:
a) How many grams of NaOH are needed to make a
30% w/w solution using deionized water as the
solvent?
Median
οƒ˜ Middle value when data are ordered from the smallest
οƒ˜ To determine: Arrange data in order, from smallest to
the highest value. If there is an even number of items
in the data set, then the median is found by taking the
mean (average) of the two middlemost numbers.
Mode
𝑀 (30𝑔)(100π‘šπ‘™)
30.0% =
30.0π‘”π‘π‘Žπ‘‚π»
𝑀
100π‘šπ‘™
οƒ˜ 30.0g would be dissolved into 70.0gH2O
☺ CONCENTRATION CALCULATIONS
οƒ˜ Changing concentrations between 2 solutions
𝐢1𝑉1 = 𝐢2𝑉2
Where:
C1 = stock concentration
C2 = concentration of new solution
V1 = volume of stock required
V2 = volume of new solution
οƒ˜ The most frequently appearing value in the data.
Measures of Dispersion
οƒ˜ Describes how widely dispersed the data in the
data set.
Range
οƒ˜ The difference between the lowest and highest
values. In {4, 6, 9, 3, 7} the lowest value is 3, and
the highest is 9, so the range is 9 βˆ’ 3 = 6. Range can
also mean all the output values of a function.
Standard Deviation
οƒ˜ Describes the spread of individual measurement
about the mean.
Example:
a) 20ml of 2.00M solution are diluted to 100ml in a
100ml volumetric flask. What is the concentration
of the new solution?
𝐢1𝑉1 (2.00𝑀)(20π‘šπ‘™)
𝐢2 =
=
= 0.4𝑀
𝑉2
100π‘šπ‘™
b) If there are 100ml of 6.00m NaOH solution but a
5.00M solution a dilute. How much water should
be added to the 6.00M solution to dilute it to
5.00M?
𝐢1𝑉1 (6.00𝑀)(100π‘šπ‘™)
𝑉2 =
=
= 120π‘šπ‘™
𝐢2
5.00𝑀
 CENTRAL TENDENCY
οƒ˜ Quantity that describes the location of a center of data.
Mean
οƒ˜ Numerical average by obtaining by dividing the sum of
the individual measurements.
οƒ˜ The mean is the average of the numbers. It is easy
to calculate: add up all the numbers, then divide by
how many numbers there are. In other words it is the
sum divided by the count.
Relative Standard Deviation
οƒ˜ Divide the standard deviation by the mean.
Percent Relative Standard Deviation.
οƒ˜ Multiply the relative standard deviation by 100%
Variance
οƒ˜ Square of the standard deviation.
 ACCURACY AND PRESCISION
Molality
Accuracy
οƒ˜ Closeness of the measurement to the TRUE or
ACCEPTED VALUE.
οƒ˜ Usually express as either absolute error or relative
error.
Absolute Error
οƒ˜ The difference between experimental result and
the true value.
οƒ˜ (x) = experimental result
οƒ˜ (µ) = true value
οƒ˜ Formula: e = x - µ
Normality
Relative Error
οƒ˜ Difference between the true value and measured
values and dividing this difference by the true value.
(ito definition sa ppt ni sir)
Temperature
Precision
οƒ˜ Measures the spread of data about a central data.
οƒ˜ Repeatability: is the precision obtained when all
measurements are made by the same analyst
during a single period of laboratory work.
οƒ˜ Reproducibility: is the precision under any other
set of condition, including that between analyst or
between laboratory sessions for a single analyst.
 OTHER EXAMPLES
Molarity
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