What volume of solution is needed to produce a 3.0 m solution of nacl containing 58.44 g of nacl?

Lab experiments and types of research often require preparation of chemical solutions in their procedure. We look at preparation of these chemical solutions by weight (w/v) and by volume (v/v). The glossary below cites definitions to know when your work calls for making these and the most accurate molar solutions.

To this we add information designed for understanding how to use the pH scale when measuring acidity or alkalinity of a solution.

Glossary, basic terms to understand...

Solute - The substance which dissolves in a solution

Solvent - The substance which dissolves another to form a solution. For example, in a sugar and water solution, water is the solvent; sugar is the solute.

Solution - A mixture of two or more pure substances. In a solution one pure substance is dissolved in another pure substance homogenously. For example, in a sugar and water solution, the solution has the same concentration throughout, ie. it is homogenous.

Mole - A fundamental unit of mass (like a "dozen" to a baker) used by chemists. This term refers to a large number of elementary particles (atoms, molecules, ions, electrons, etc) of any substance. 1 mole is 6.02 x 1023 molecules of that substance. (Avogadro's number).M

Introduction to preparation of solutions.

Many experiments involving chemicals call for their use in solution form. That is, two or more substances are mixed together in known quantities. This may involve weighing a precise amount of dry material or measuring a precise amount of liquid. Preparing solutions accurately will improve an experiment's safety and chances for success.

Solution 1: Using percentage by weight (w/v)

Formula

The formula for weight percent (w/v) is: [Mass of solute (g) / Volume of solution (ml)] x 100

Example

A 10% NaCl solution has ten grams of sodium chloride dissolved in 100 ml of solution.

Procedure

Weigh 10g of sodium chloride. Pour it into a graduated cylinder or volumetric flask containing about 80ml of water. Once the sodium chloride has dissolved completely (swirl the flask gently if necessary), add water to bring the volume up to the final 100 ml. Caution: Do not simply measure 100ml of water and add 10g of sodium chloride. This will introduce error because adding the solid will change the final volume of the solution and throw off the final percentage.

Solution 2: Using percentage by volume (v/v)

When the solute is a liquid, it is sometimes convenient to express the solution concentration as a volume percent.

Formula

The formula for volume percent (v/v) is: [Volume of solute (ml) / Volume of solution (ml)] x 100

Example

Make 1000ml of a 5% by volume solution of ethylene glycol in water.

Procedure

First, express the percent of solute as a decimal: 5% = 0.05

Multiply this decimal by the total volume: 0.05 x 1000ml = 50ml (ethylene glycol needed).

Subtract the volume of solute (ethylene glycol) from the total solution volume:

1000ml (total solution volume) - 50ml (ethylene glycol volume) = 950ml (water needed)

Dissolve 50ml ethylene glycol in a little less than 950ml of water. Now bring final volume of solution up to 1000ml with the addition of more water. (This eliminates any error because the final volume of the solution may not equal the calculated sum of the individual components).

So, 50ml ethylene glycol / 1000ml solution x100 = 5% (v/v) ethylene glycol solution.

Solution 3: Molar Solutions

Molar solutions are the most useful in chemical reaction calculations because they directly relate the moles of solute to the volume of solution.

Formula

The formula for molarity (M) is: moles of solute / 1 liter of solution or gram-molecular masses of solute / 1 liter of solution.

Examples

The molecular weight of a sodium chloride molecule (NaCl) is 58.44, so one gram-molecular mass (=1 mole) is 58.44 g. We know this by looking at the periodic table. The atomic mass (or weight) of Na is 22.99, the atomic mass of Cl is 35.45, so 22.99 + 35.45 = 58.44.

If you dissolve 58.44g of NaCl in a final volume of 1 liter, you have made a 1M NaCl solution, a 1 molar solution.

Procedure

To make molar NaCl solutions of other concentrations dilute the mass of salt to 1000ml of solution as follows:

0.1M NaCl solution requires 0.1 x 58.44 g of NaCl = 5.844g

0.5M NaCl solution requires 0.5 x 58.44 g of NaCl = 29.22g

2M NaCl solution requires 2.0 x 58.44 g of NaCl = 116.88g

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What is molarity?

Molarity is the concentration of a solution in terms of the number of moles of the solute in 1 dm3 (1 liter) of the solution.

The units of molarity are M or mol/L. A 1 M solution is said to be “one molar”.

A mole is the quantity of anything that has the same number of particles as 12 g of carbon-12. This equates to roughly 6.02x1023, also referred to as Avogadro's Number. So, 1 mole of hydrogen gas (H2) contains 6.02x1023 molecules, and 1 mole of glucose (C6H12O6.) also contains 6.02x1023 molecules, but as H2 is a much simpler molecule, 1 mole of H2 will have a much smaller mass (the molar mass) than 1 mole of C6H12O6.

The molar mass is the mass in grams of 1 mole of a particular molecule.

One mole of sodium (Na) is 22.99 g, and 1 mole of chlorine is 35.45 g. For sodium chloride (NaCl) they are in a ratio of 1:1 so the molar mass of NaCl is 22.99 + 35.45 = 58.44 g/mol.
For a compound like water (H2O), 1 mole of hydrogen (H) is 1.008 g/mol and 1 mole of oxygen (O) is 15.9994 g/mol. So, the molar mass of H2O is (2 x 1.008) + 15.9994 = 18.0154 g/mol.

Two solutions that have the same molarity will have the same number of molecules of the chemical per liter but are likely to contain differing masses of that chemical per liter to achieve this. Whereas two solutions at the same concentration will have the same mass of the chemical per liter of solution but are therefore likely to have differing numbers of molecules of that chemical per liter. Provided some additional information is known, one value can be deduced from the other using the equations below.

To calculate molarity or to calculate related values (including volume, mass, molar mass and concentration) from molarity, the following equations are utilized.

Number of moles (mol) = Mass (g) / Molar Mass (g/mol) Concentration (g/L) = Mass (g) / Volume (L)

Molarity (M or mol/L) = Number of Moles (mol) / Volume (L)

The scenario in which most lab scientists will encounter this type of calculation is when making up solutions following a standard operating procedure (SOP) or a scientific paper. Here, the solution being used is typically defined by its molar concentration (M). For example;
You need to make a 0.5 M solution of NaCl, having decided you want 2 liters how much NaCl should you add?

1. First you must calculate the number of moles in this solution, by rearranging the equation
   No. Moles (mol) = Molarity (M) x Volume (L) = 0.5 x 2 = 1 mol
2. For NaCl, the molar mass is 58.44 g/mol Now we can use the rearranged equation Mass (g) = No. Moles (mol) x Molar Mass (g/mol) = 1 x 58.44 = 58.44 g

   So, to make 2 liters of a 0.5 M solution of NaCl, you would need to add 58.44 g of NaCl.

   
   

– mass of one atom of that element which typically reflects the mass of the nucleus (protons plus neutrons). For example, hydrogen is 1. This used to be measured in atomic mass units (AMU) but is now typically expressed in Daltons (Da).

- sum of the atomic weights of all atoms appearing in a given molecular formula. For example, glucose has a molecular formula of C6H12O6, the molecular weight of C is 12 Da, H is 1 Da and O is 16 Da. Therefore, the molecular mass of glucose is = (6 x 12) + (12 x 1) + (6 x 16)

– values for atomic and therefore molecular masses are normally obtained relative to the mass of the isotope 12C (carbon-12), however “relative” is generally omitted from the title. Written correctly, the relative values have no units.

- sum of the atomic weights of all atoms appearing in a given empirical formula. The empirical formula indicates the ratio of atoms of each element in a molecule rather than the actual number. For example, glucose (molecular formula C6H12O6) would therefore have the empirical formula CH2O, the molecular weight of C is 12 g, H is 1 g and O is 16 g. Therefore, the formula mass of glucose is

For molecules like H2O, where the formula is already in its simplest form, the formula mass and molecular mass are the same.

– similar to molarity, however calculation of normality uses the number of mole equivalents rather than the number of moles. Units are N or eq/L. Normality is generally only used when a substance has more than one sub-species that can participate in a specified reaction such as a proton for acid/base reactions, an electron for oxidation/reduction reactions, or in precipitation reactions. For example, sulfuric acid (H2SO4) has two ionizable protons (H+) which can participate in the neutralization of a base such as sodium hydroxide (NaOH)

– similar to molarity, however calculation of molality uses the mass rather than volume of the solvent used, making it temperature independent unlike molarity. Units are m or mol/kg.