Welcome to the World of MOLARITY
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sample problems given.
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MOLARITY
Molarity expresses the concentration of a solution
in number of moles of solute per liter of solution.
Mathematically;
M=n/V
where M=molarity (in moles solute/liter solution)
n=number of moles of solute
V=volume of solution (in liters)
SAMPLE PROBLEM # 1
A solution of sodium hydroxide was prepared by dissolving
10.0 grams of sodium hydroxide (NaOH) in distilled water
making the volume up to 500 cm3. What is the molarity of
the NaOH solution? Atomic masses: Na=23; O=16, H=1
GIVEN: mass of solute (NaOH)= 10.0 grams
molar mass of NaOH=1(23)+1(16)+1(1) =40g/mole
n (moles of NaOH)= 10.0g(1mol/40g)=0.25moles
V of NaOH solution= 500cm3(1L/1000cm3)=0.50L
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Solution to SP # 1
GIVEN: mass of solute (NaOH)= 10.0 grams
molar mass of NaOH=1(23)+1(16)+1(1) =40g/mole
n (moles of NaOH)= 10.0g(1mol/40g)=0.25moles
V of NaOH solution= 500 cm3(1L/1000cm3)=0.50L
SOLUTION:
M(NaOH solution)=n(moles NaOH)/V(volume solution in L)
=0.25moles /0.50L
= 0.5 M or 0.5 moles/liter
A 0.5 M solution, means
there are 0.5 moles of NaOH in 1 liter of NaOH solution
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SAMPLE PROBLEM # 2
Calculate the number of moles copper sulfate
in a 250 milliliter (mL) of 0.1 molar copper
sulfate (CuSO4) solution.
GIVEN: M =0.1 moles/L
V of CuSO4 solution= 250mL(1L/1000mL)=0.25L
n (moles of CuSO4)= ?
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Solution to SP # 2
GIVEN: M =0.1 moles/L
V of CuSO4 solution= 250mL(1L/1000mL)=0.25L
n (moles of CuSO4)= ?
SOLUTION:
n(moles CuSO4)=(V of CuSO4 solution in L)x (M of CuSO4 solution)
=(0.25 L) (0.1 moles/ L)
= 2.5 moles CuSO4
There are 2.5 moles of CuSO4 in a 250 mL
of 0.1 M CuSO4 solution
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SAMPLE PROBLEM # 3
1. What is the volume of a 0.50 molar
potassium bromide (KBr) solution
containing 1.5 moles of potassium bromide
(KBr)?
GIVEN: M =0.5 moles/L means there are 0.5 moles KBr in
1L KBr solution
n =1.5 moles KBr
V of KBr solution=?
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Solution to SP # 3
GIVEN: M =0.5 moles/L means there are 0.5 moles KBr in
1L KBr solution
n =1.5 moles KBr
V of KBr solution=?
SOLUTION:
V of KBr solution in L) =n(moles KBr)/( M of KBr solution)
= (1.5 moles)
(0.5 moles/ L)
= 3.0 L KBr
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SAMPLE PROBLEM # 4
3. A 250 milliliter bottle of vinegar reads 5% acidity (5 g
acetic acid in 100 gram vinegar).Calculate the
concentration of acetic acid in moles per liter. (1mole
CH3COOH=60 grams)
GIVEN: V of solution =250 mL
n acetic acid=250mL (5g acetic acid)(1mole acetic acid)
(100mL)
(60g acetic acid)
=0.208 moles acetic acid
M=0.208 moles
0.25 L
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M=0.83 moles/liter
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THINK PAIR
SHARE
5. Write your
comments on
each output
posted.
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1. Find your
pair
.
2.Discuss how
you are going to
solve the
solutions at work
problems
4. Post your
outputs in the
“output bulletin”
provided in our
classroom.
3. Write your
answers in
“SOLUTIONS AT
WORK” SHEET
#1
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SOLUTIONS AT WORK Sheet #1
1. A solution of glucose (C6H12O6) contains 300 grams in
1.0 liter of solution. What is the concentration of the
glucose solution in moles per liter? Atomic masses:
C=12, H=1, O=16
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SOLUTIONS AT WORK Sheet #1
2. A 3.0 grams of magnesium hydroxide, Mg(OH)2 is
dissolved in water to produce a 0.2 molar magnesium
hydroxide solution. What is the volume of the solution
prepared? Atomic masses: Mg=24, H=1, O=16
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Assignment: Answer the following briefly:
There are times when glucose solution must be
introduced into the bodies of patient. Why is it
dangerous for this solution to be highly
concentrated?
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