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)