1. B- The atomic weights are: Na (sodium) = 23, and Cl
(chloride) = 35.5
23+35.5 = 58.5
2. D – The atomic weights are: H (hydrogen) = 1 (x 2) since you have 2
molecules, S (sulfur)=32, and O (oxygen) = 16 (x 4) since you have 4 molecules
of oxygen.
2+32+ 64 = 98
3. E – The atomic weights are: Ca (calcium)=40, Cl(chloride)= 35.5 x
2, H= 1 x 2, O = 16
40+71+2+16= 129
4. A – The atomic weights are: H = 1, Cl=35.5
1+35.5 = 36.5
5. C – The atomic weights are: H=1, N (nitrogen) = 14, O (oxygen) = 16
x 3
1+14+48 = 63
6. A – 50 milliliters; this is solved using the formula, C1V1=C2V2
Where: C1-
concentration 1 (original concentration)
V1 – volume 1 (original volume)
C2 – concentration 1 (final
concentration)
V2 – volume 2 (final
volume)
Analyzing the
problem, you can substitute the values now.
(0.5)V1) =
(0.1)(250)
V1 =
(0.1)(250)/0.5
V1 = 50
milliliters of 0.5 N HCl
7. D – 1.275 milliliters or 1.3; you can solve it using two methods.
The first method is through ratio and proportion. Normal Saline Solution (NSS)
is 0.85% NaCl; it is the solute. If you want to round it off, to 0.9% you can.
But for this particular problem, we will use 0.85%. The 0.85% NaCl is 0.85
grams of NaCl diluted to 100 milliliters of distilled water up to the 100 mark
of a volumetric flask. Since 0.85 grams is to 100, we’ll make this as the
basis:
0.85/100 = x/150
Transposing the equation to get x, we come up with this:
X= 0.85 x 150/100
X = 1.275 or 1.3 milliliters
The second method is to use the percent formula, which is:
%=Weight in grams/Total volume of solution X 100.
Substituting the values:
85= W in g/150 x 100
Transpose to isolate W in g and you’ll come up with this
formula:
W in g = 0.85 x 150/100 = 1.275 grams of NaCl
Tips to remember:
* When solving for percent solutions in relation to weight and
volume, convert first the volume to milliliters if it is not in milliliters.
* If the volume is 100 milliliters, substitute the percent sign
with the grams and that is already the weight of the solute.
Examples:
·
5% CaCl2 is actually
prepared by weighing 5 grams of CaCl2 and dilute it with distilled water up to
the 100 mark in a volumetric flask. The volumetric flask is the most accurate
glassware in the preparation of solutions.
·
10% NaOH = 10 grams
NaOH diluted with distilled water up to the 100 mark in a volumetric flask.
9. C – 222 grams; to determine the weight of a substance when
preparing for a Molar solution, use the formula:
W= DV x DM x MW
Where:
W – Weight in grams
DV- Desired
Volume (in liters)
DM – Desired Molarity
MW – Molecular Weight
Substituting
the values, we get:
W= 1 liter x 2
M x 111
W = 222 grams
of CaCl2
10. A – 1.37 N; You can use the formula for Normality, which is N = (Volume
of solute/Volume of solution/MW/f)/Volume of solution in liters, and the
formula for percent (V/V), which is % = Volume of solute in milliliters/Total
volume of solution in milliliters x 100.
With the first formula, the substitution would be:
N=
((0.5/100)/(36.5/1)/1
N =
(0.005)/36.5
N = 1.3698 0r
1.37
The volume of solution is assumed to be 100 milliliters because
it is a percent solution.
You can also use the conversion units, making use of the percent
formula. For Normality your units would be milliequivalent per liter of
solution. (MEq/L)
