# Solution: Grams Essay

Submitted By aisha271005
Words: 721
Pages: 3

Examples: Making Solutions

Two simple examples are presented here. A third example is of a complex solution for which the description lists the concentrations of components using different expressions.

Weight in volume: Prepare 2 liters 0.85% sodium chloride
With 1% defined as 1 gram per 100 ml, 0.85% is 0.85 grams per 100 ml. Since two liters is 20x the volume of 100 ml, we need 20 x 0.85 grams which is 17 grams NaCl. For this quantity we can use a top loading balance or even a trip balance.
A typical electronic balance is accurate to one hundredth of a gram, which is sufficiently accurate for weighing out 17 grams. First we "tare" the instrument by placing a weigh boat onto the pan and setting it to “zero.” We don't want to contaminate our chemical stocks, so we either clean the spatula or spoon before dipping it into the container or we simply shake the chemical out onto the boat.

Suppose that we tap out 16.97 grams of NaCl. Should we go to the trouble to get that last 0.03 gram? Nope! Consider that if it was necessary to be more accurate, we would describe the formula as something like 0.846% NaCl, or maybe 0.8495%. If there is some advantage to being precise then we should exercise precision, otherwise trying to be too precise just wastes time.

Remember how to use significant digits? Seventeen grams means greater than 16.5 grams and less than 17.5 grams. If we wanted to be more accurate we would write “17.0” grams, meaning greater than or equal to 16.95 grams and less than or equal to 17.05 grams.

Molarity: Prepare 200 ml of 70 mM sucrose
Suppose that you need 200 milliters of a 70 mM solution of sucrose. Two hundred milliliters is 0.2L and 70 mM is 0.07M. The molecular weight of sucrose can be determined from its chemical formula, namely C12H22O11 and the atomic weights of carbon, hydrogen, and oxygen. The formula weight for sucrose is identical to its molecular weight, namely 342.3 grams per mole. A 1M solution would consist of 342.3 grams sucrose in one liter final volume.

A concentration of 70 mM is the same as 0.07 moles per liter. Take 0.07 moles/liter times 342.3 grams per mole and you have 23.96 grams needed per liter. To make 200 milliliters of your solution multiply grams/liter by liters needed. Since 200 milliliters is 0.2L, multiply 23.96 grams by 0.2L to get 4.792 grams needed. Since a typical top loading electronic balance displays mass to the nearest 0.01 gram, the amount to be weighed should be rounded to 4.79 grams, although it is perfectly acceptable and perhaps even preferable to round to 4.8 grams.

Complex solution: Prepare a sample buffer for SDS-PAGE

The following formula describes the composition of the 2x concentrated buffer that we use to denature proteins for electrophoresis. The formula descriptions v/v or w/w would not be listed in a methods section since it is obvious which…