Answers to the sample problems.

1. WORKSHEET 1:
LABORATORY REAGENTS PREPARATION AND
CALCULATIONS
1. How can the following solutions be prepared?
a. 100 mL of 40% (w/v) polyethylene glycol (PEG)
% w/v = Mass of solute x 100
Vol. of solution
40% = 0.4 g/ml
0.4 g/ml = x g
100 ml
x g = (0.4 g/ml) (100) = 40 g
Weigh out 40 g of PEG and
dissolve in distilled water
so that the final volume of
the solution, with the PEG
completely dissolved, is
100 mL.

2. WORKSHEET 1:
LABORATORY REAGENTS PREPARATION AND
CALCULATIONS
b. 47 mL of 7% (w/v) solution of sodium chloride (NaCl)
% w/v = Mass of solute x 100
Vol. of solution
7 % = 0.07 g/ml
0.07 g/ml = x g
47 ml
x g = (0.07 g/ml) (47 ml) = 3.29 g
Weigh out 3.29 g of NaCl and dissolve
the crystals in some volume of
distilled water less than 47 mL, a
volume measured so that, when the
3.29 g of NaCl are added, it does not
exceed 47 mL.
When the NaCl is completely
dissolved, dispense the solution into
a 50 mL graduated cylinder and bring
the final volume up to 47 mL with
distilled water.

3. WORKSHEET 1:
LABORATORY REAGENTS PREPARATION AND
CALCULATIONS
c. 200 mL of 95% (v/v) solution of ethanol
% v/v = Vol. of solute x 100
Vol. of solution
95 % = 0.95
0.95 = x ml
200 ml
x ml = (0.95) (200 ml) = 190 ml
Measure 190 mL of 100% (200 proof) ethanol and add 10 mL of distilled water
to bring the final volume to 200 mL.
OR
Ethanol : Dist. H2O
95 % 0.05 %
0.95 0.05
(0.95) (200 ml) (0.05) (200 ml)
190 ml 10 ml

4. 2. If 25g of NaCl is dissolved into a final volume of 500mL, what is the percent
(w/v) concentration of NaCl in the solution?
% w/v = Mass of solute x 100
Vol. of solution
% w/v = 25 g NaCl x 100
500 ml
% w/v = 5 % NaCl
OR
We can set up an equation of two ratios in which x
represents the unknown number of grams. This
relationship is read ‘x g is to 100 mL as 25 g is to
500 mL.
x g = 25g
100 mL 500 mL
Solving for x gives the following result:
x g (25g)(100 mL) = 2500g = 5 g
500 mL 500 mL
Therefore, there are 5 g of NaCl in 100 mL (5
g/100 mL), which is equivalent to a 5% solution.

5. 3. How will you prepare 200mL of 0.3M NaCl?
OR
Get first the wt. (g) of NaCl needed
for 1 L of a 0.3 M solution.
58.45 g = x g
1 M 0.3 M
g NaCl = 17.53 g
Therefore, 17.53 g per Liter
17.53 g = x g
1000 mL 200 mL
x g = 3.51
Using this formula,
gs = MMs x M x V
MMs of NaCl = 58.45 g/mol
M = 0.3 M = 0.3 mol/L
V = 200 mL = 0.2 L
gNaCl = 58.45 g x 0.3 mol x 0.2 L
mol L
= 3.51 g
Therefore, to prepare 200 mL of 0.3 M NaCl solution, 3.51 g of NaCl are dissolved
in distilled water to a final volume of 200 mL.

6. 4. From a stock solution of 1M Tris solution, how will you prepare 400mL of 0.2M?
Therefore, to 320 mL (400 mL - 80 mL = 320 mL)
of distilled water, add 80 mL of 1 M Tris, pH 8.0,
to bring the solution to 0.2 M and a final
volume of 400 mL.
Using this formula,
M1V1 = M2V2
(1 M) (V1) = (0.2 M) (400 mL)
V1 = (0.2 M) (400 mL)
1 M
= 80 ml

7. 5. How is 4mL of 50mM NaCl solution prepared from a 2M NaCl stock?
Therefore, to 3.9 mL of distilled water, add 0.1 mL of 2 M NaCl stock solution to produce 4
mL final volume of 50 mM NaCl.
First, a conversion factor must be included in the
equation so that molarity (M) can be converted to
millimolarity (mM).
2 M x 1000 mM = 2000 mM
1 M
Then, using this formula,
M1V1 = M2V2
(2000 mM) (V1) = (50 mM) (4 mL)
V1 = (50 mM) (4 mL)
2000 mM
= 0.1 ml

8. 6. What is the molar concentration of a 10% NaCl solution?
Therefore, a 10% NaCl solution is equivalent to 1.86 M NaCl.
At 25⁰C, the density of 10% NaCl is 1.09 g/ml
Mass of solution = vol x density
= 1,000ml x 1.09 g/mL = 1,090 g
Mass % = 10 % = 0.10
Molar mass of NaCl = 58.45 g
To get molarity of 10% NaCl:
(Mass of sol’n)(Mass%) = 1,090g x 0.10 = 1.86 moles
MMNaCl 58.45g / mol
1 M = moles x 1.86 moles = 1.86 M
Liter

9. OR We can use equations of ratios (if density value is not given.)
Therefore, a 10% NaCl solution is equivalent to 1.71 M NaCl.
Note: the value is close to the first computation
First, 10 % NaCl contains 10 g of NaCl in 100 mL H2O. Determine the mass of NaCl in 1000 mL.
10 g = x
100mL 1000 mL
x = 100 g ---> 1000 mL solution of 10% NaCl contains 100 g of NaCl.
Then, using the gram molecular weight of NaCl (58.45), an equation of ratios can be written
to determine molarity. In the following equation, we determine the molarity (M) equivalent
to 100 g:
x M = 1 M
100 g 58.45 g
x = 1 M (100 g)
58.45 g
= 1.71 M NaCl