Prelab 1 -BIO3126
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Prelab 1-0
Exercise 1.1 1.
What is the molarity of the methylene blue stock solution? (1 point)
0.26%
m
v
of methylene
¿
=
x gsolute
100
mL solvent
x
=
0.26
gof solute
0.26
gof solute ×
1
mole
320
g
=
8.215
×
10
−
4
mol
8.215
×
10
−
4
mol÷
0.100
L
=
8.215
×
10
−
3
M of methylene
¿
stock
Therefore, molarity of methylene blue stock solution is 8.215
×
10
−
3
M
~ 8.215 mM
2.
What dilution is required to prepare 5 mL of a 0.4mM starting solution? (1 point)
C
1
V
1
=
C
2
V
2
8.215
mM ×V
1
=
0.4
mM ×
5.0
m L
V
1
=
0.2462
mL≈
0.25
mL
5
mL solution
=
0.2462
mL
+
water
water
=
5
mL
−
0.2462
mL
=
4.7538
mL≈
4.75
mL
Therefore the dilution is 0.25:5 which means 1/20 dilution 3.
What volumes of the methylene blue stock and water are required to prepare the above-
described solution? (1 point)
0.2462
mL
≈
0.25
mL
for stock solution of methylene blue and 4.75 mL of water. 4.
Complete the following table: (2 points)
Tube
Number
Source tube
number
Vol. of
water (mL)
Vol. from
source tube
(mL)
Dilution
factor (X)
Serial
dilution
factor (X)
2
1
1.2
4.8
1.25
1.25
3
2
3.0
2.5
2.2
2.75
4
3
1.5
2.0
1.75
4.81
5
4
1.0
1.6
1.63
7.82
6
5
1.0
1.5
1.67
13.06
Note: Dilutions must be expressed as a fraction where the numerator is 1. Ex. 1/5. Dilution factors must be expressed as a number followed by the symbol “X”. Ex. 5X.
Exercise 1.2
5.
What weight of sugar and salt are required to prepare 30 mL 30% solutions of each solute? (1 point)
30%
=
x g
30
mL
×
100
x
=
9.0
g
9.0 g of sugar and 9.0 g of salt are required to make a 30% (m/v) solution 6.
What are the percent concentrations of glucose and sodium chloride in TSB? (1 point)
Glucose concentration: 0.25%
=
2.5
g
1000
mL
×
100
Salt concentration:
0.5%
=
5.0
g
1000
mL
×
100
Total salt and glucose concentration = 0.25% + 0.5% = 0.75% (m/v)
7.
How many grams of glucose and sodium chloride are present in 5 mL of 2X TSB? (1 point)
There are 2.5 g of glucose and 5.0 g of sodium chloride in 1000 mL TSB, so in 1000 mL of 2X TSB, there will be 5 g of glucose and 10 g of sodium chloride. Grams of glucose:
5
g
1000
mL
=
x g
5.0
mL
25
g
mL
=
1000
x
g
mL
x
=
0.025
g
Grams of sodium chloride:
10
g
1000
mL
=
x g
5.0
mL
50
g
mL
=
1000
x
g
mL
x
=
0.05
g
Therefore, in 5 mL of 2X TSB, there is 0.05 g of NaCl and 0.025 g of glucose. 8.
Complete the following table: (1 point)
Sugar conc. (%)
Vol. stock
soln. (mL)
Vol. of
water (mL)
Vol. 2X TSB
(mL)
Dilution
factor (X)
0.5
0.083
4.917
5.0
60
5
1.583
3.417
5.0
3.16
10
3.250
1.750
5.0
1.54
15
4.917
0.083
5.0
1
Salt conc. (%)
0.5
0.000
5.000
5.0
0
5
1.500
3.500
5.0
3.33
10
3.167
1.833
5.0
1.58
15
4.833
0.167
5.0
1
Sample calculation for sugar conc. (%) at 0.5%
0.05%
=
x
10
mL
×
100
x
=
0.05
g
0.05 g of sugar in the final solution.
Since there is 0.025 g of glucose in the 5 mL of 2X TSB, then,
0.05
g
−
0.025
g
=
0.025
g
There should be 0.025 g of sugar is needed from the 30 mL stock solution
Since we have 9.0 g of sugar needed to make the stock solution (30% m/v) of 30 mL,
9
g
30
mL
=
0.025
g
x
x
=
0.083
mL
Vol. of water (mL): 10 mL – 0.05 mL – 0.083 mL = 4.917 mL Therefore, the vol. of solution is 0.083 mL and the volume of water is 4.917 mL.
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F3
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F5
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F6
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J
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0:50-
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on 12 of 21 >
Macmillan Learning
1
Q
A solution contains 0.575 g Ca²+ in enough water to make a 1825 mL solution. What is the milliequivalents of Ca²+ per liter
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concentration:
2
3
E
80
$
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R
888
F4
%
5
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MacBook Air
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