Answers to Question on Laboratory Mathematics

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)