# CHEM-REVIEWER 2nd sem prelims ```CHEM REVIEWER
☺ MOLARITY
unit: M or mol/L
 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
 𝑀=
49𝑔
= 1.0𝑚𝑜𝑙/𝑘𝑔
(98 𝑔/𝑚𝑜𝑙)(0.5𝑘𝑔)
𝑚=
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
since 𝑚𝑜𝑙𝑒𝑠 = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡
𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
(𝑀𝑊 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒)(𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒)
Example:
1. Calculate the molarity of 2.0 L of solution containing
5.0 moles of NaOH.
𝑀=
5.0𝑚𝑜𝑙 𝑠𝑜𝑙𝑢𝑡𝑒
= 2.5𝑀 𝑜𝑟 2.5𝑚𝑜𝑙/𝐿
2.0𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
☺ SOLUTIONS
 Homogenous mixtures of 2 or more substances
 Parts:
1) Solute – substance that dissolves in a solution
2) Solvent – dissolving medium
☺ MOLE FRACTION
 The number of moles of 1 component to the total
number of moles of all components present
𝑀𝑜𝑙𝑒 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑒 =
𝑚𝑜𝑙𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 + 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑀𝑜𝑙𝑒 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑜𝑙𝑣𝑒𝑛𝑡 =
𝑚𝑜𝑙𝑒 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 + 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
2. Calculate the volume of a 0.750M sulfuric acid
(H2SO4) solution containing 0.120 moles of solute.
0.120𝑚𝑜𝑙
0.750𝑀 =
𝑥𝐿
𝑥 = 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:
 10g sulfuric acid
10𝑔
𝑀 = (98)(0.5𝐿) = 0.204𝑚𝑜𝑙/𝐿
 98 MW
 500mL
𝑚𝑜𝑙𝑒𝑠 𝐻2𝑂 =
2. Calculate the weight in grams of sulfuric acid in 2L of
0.100M solution
Given:
 0.100M solution
𝑔𝑟𝑎𝑚𝑠 = (𝑀)(𝑀𝑊)(𝐿)
 2L
𝑔𝑟𝑎𝑚𝑠 = (0.100)(98)(2𝐿)
 98 MW
𝑔𝑟𝑎𝑚𝑠 = 19.6𝑔
☺ MOLALITY
𝑚=
1. What are the mole fractions of solute and solvent in a
solution prepared by dissolving 98g sulfuric acid in
162 water?
98𝑔
𝑚𝑜𝑙𝑒𝑠 𝐻2𝑆𝑂4 =
= 1𝑚𝑜𝑙 (𝑠𝑜𝑙𝑢𝑡𝑒)
98𝑔/𝑚𝑜𝑙
unit: m or mpl/kg
𝑚𝑜𝑙 𝑠𝑜𝑙𝑢𝑡𝑒
(𝑀𝑊)(𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡)
Example:
1. What is the molality of a solution in which 49g sulfuric
acid is dissolved in 500g H20?
162𝑔
= 9 𝑚𝑜𝑙 (𝑠𝑜𝑙𝑣𝑒𝑛𝑡)
18𝑔/𝑚𝑜𝑙
𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝐻2𝑆𝑂4 =
1
= 0.1
1+9
𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝐻2𝑆𝑂4 =
9
= 0.9
1+9
☺ NORMALITY FRACTION
 Gram equivalent weight of solute per liter solution
Gram equivalent =
𝑁=
MW
𝑣𝑎𝑙𝑒𝑛𝑐𝑒
𝑔𝑟𝑎𝑚𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
(𝐸𝑊)(𝐿𝑖𝑡𝑒𝑟𝑠 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛)
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 𝑜𝑟 𝑒𝑞𝑢𝑖𝑣/𝐿
98
∗
2𝐿
2
☺ PERCENT SOLUTION
 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
 Volume = 100
𝑤
𝑔𝑟𝑎𝑚𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
% =
𝑥 100
𝑣 𝑚𝑙 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
 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?
%
𝑤
25.0𝑔 𝑁𝑎𝐶𝑙
=
𝑥 100 = 25%
𝑣 100𝑚𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
 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 𝑁𝑎𝑂𝐻 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑠𝑢𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑡𝑜 200𝑚𝑙
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 𝑚𝑙 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
=
𝑥 100
𝑣
100𝑚𝑙
 Or ml of solute per 100ml of solution
 % v/v is similar to % w/w in that the total volume of
the solution is 100ml
 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 100𝑚𝑙 𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 −
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
Example:
a) Prepare 200ml of 5% ammonium hydroxide
solution
𝑣
𝑚𝑙 𝑠𝑜𝑙𝑢𝑡𝑒
% =
𝑥 100
𝑣 100𝑚𝑙 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑚𝑙 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 =
𝑥=
%𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑥 𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑒𝑠𝑖𝑟𝑒𝑑
100
(5)(200)
= 10𝑚𝑙
100
 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?
𝑥 = 100 − 75
𝑥 = 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)
%
𝑤 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
=
𝑥 100
𝑤 100𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
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)
𝑥=
= 39.7𝑚𝑙
100
 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
 Middle value when data are ordered from the smallest
 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
𝑤 (30𝑔)(100𝑚𝑙)
30.0% =
30.0𝑔𝑁𝑎𝑂𝐻
𝑤
100𝑚𝑙
 30.0g would be dissolved into 70.0gH2O
☺ CONCENTRATION CALCULATIONS
 Changing concentrations between 2 solutions
𝐶1𝑉1 = 𝐶2𝑉2
Where:
C1 = stock concentration
C2 = concentration of new solution
V1 = volume of stock required
V2 = volume of new solution
 The most frequently appearing value in the data.
Measures of Dispersion
 Describes how widely dispersed the data in the
data set.
Range
 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 − 3 = 6. Range can
also mean all the output values of a function.
Standard Deviation
 Describes the spread of individual measurement
Example:
a) 20ml of 2.00M solution are diluted to 100ml in a
100ml volumetric flask. What is the concentration
of the new solution?
𝐶1𝑉1 (2.00𝑀)(20𝑚𝑙)
𝐶2 =
=
= 0.4𝑀
𝑉2
100𝑚𝑙
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?
𝐶1𝑉1 (6.00𝑀)(100𝑚𝑙)
𝑉2 =
=
= 120𝑚𝑙
𝐶2
5.00𝑀
 CENTRAL TENDENCY
 Quantity that describes the location of a center of data.
Mean
 Numerical average by obtaining by dividing the sum of
the individual measurements.
 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
 Divide the standard deviation by the mean.
Percent Relative Standard Deviation.
 Multiply the relative standard deviation by 100%
Variance
 Square of the standard deviation.
 ACCURACY AND PRESCISION
Molality
Accuracy
 Closeness of the measurement to the TRUE or
ACCEPTED VALUE.
 Usually express as either absolute error or relative
error.
Absolute Error
 The difference between experimental result and
the true value.
 (x) = experimental result
 (&micro;) = true value
 Formula: e = x - &micro;
Normality
Relative Error
 Difference between the true value and measured
values and dividing this difference by the true value.
(ito definition sa ppt ni sir)
Temperature
Precision