Question

What volume of glycerol,\(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}\) \((d=1.26 \mathrm{g} / \mathrm{mL})\)must be added per kilogram of water to produce a solution with 4.85 mol \% glycerol?

Short answer

To prepare a solution with 4.85 mol % glycerol by adding it to a kilogram of water, approximately 62.4 mL of glycerol is required.

Step-by-step solution

Determine the Molar Mass of Glycerol

The molar mass of glycerol, which has the chemical formula \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{CH}_{2} \mathrm{OH}\), is found to be \(92.09 \mathrm{g/mol}\); this can be calculated by adding up the molar masses of individual atoms.

Calculation of Moles in 4.85% of 1kg of water

First, calculate the total moles in solution when 4.85 mol % of glycerol is added to 1 kg (or 1000 grams) of water. We know that mol percent (mol%) is defined by the number of moles of the solute (glycerol) per 100 moles of the total solution. Therefore, if 'x' moles of glycerol constitute 4.85% by mole, then the total moles in the solution will be \(x/0.0485\). The moles of water can be found by the formula weight of water (1kg) divided by the molar mass of water (18.015 \(g/mol\)), which equals \(1000/18.015\) moles. To solve for 'x', we set up the equation \(x/0.0485 = x + 1000/18.015\) then solve for 'x'.

Determine the Mass of Glycerol

The mass of glycerol can be found by multiplying the moles of glycerol 'x' (found in step 2) by the molar mass of glycerol.

Determine the Volume of Glycerol

The volume of glycerol can be found by using the density formula, Density = Mass/Volume. We can rearrange this to Volume = Mass/Density. therefore we find the volume of glycerol by dividing the mass of glycerol (found in step 3) by the given density of glycerol (1.26 \(g/mL\)).

Key concepts

Density

Density is a crucial concept when it comes to solving problems related to finding volume from mass in chemistry. Density is defined as mass per unit volume. It tells us how much mass is compacted within a given space. In this exercise, glycerol is the substance in question, and its density is given as 1.26 grams per milliliter. This means that for every milliliter of glycerol, there is 1.26 grams of mass present.

To calculate the volume of glycerol, we can rearrange the density formula. The formula for density is given by:

• Density = Mass/Volume

To find the volume, you can rearrange it to:

• Volume = Mass/Density

This rearranged formula allows us to calculate the volume if we know the mass and density. The density concept is helpful to determine the space that a substance will occupy when you have its mass.

Molar Mass

The molar mass is another fundamental idea in chemistry, and it describes the mass of one mole of a substance. For glycerol, the molar mass is calculated by summing up the atomic masses of all atoms in its chemical formula - \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}\).Knowing the molar mass is important as it acts as a conversion factor between grams and moles.

In the given problem, the molar mass of glycerol is determined to be 92.09 grams per mole. To find this:

• Add the atomic masses of all carbon (C), hydrogen (H), and oxygen (O) atoms in the structure.

The periodic table provides these atomic masses, typically:

• Carbon: approximately 12.01 g/mol
• Hydrogen: approximately 1.01 g/mol
• Oxygen: approximately 16.00 g/mol

Understanding molar mass helps in balancing reactions, finding reactant amounts, and determining how much of a substance is needed to react completely in a given situation.

Solution Concentration

Solution concentration is a measure of the amount of solute present in a given quantity of solvent or solution. It is often expressed as a percent by mass, volume, or in this case, a percentage by mole. When dealing with molecular quantities and reactions, having the correct concentration ensures that reactions occur properly.

In the exercise, the concentration of glycerol is given as 4.85 mol %. This means that for every 100 moles of the total solution, 4.85 moles are contributed by glycerol. To execute such calculations:

• Calculate the total moles needed in the solution.
• Use the molar mass to find the mass of glycerol required.

This helps in determining the exact composition of the mixture, ensuring that the proportion by moles of the solute (glycerol) is accurate relative to the whole solution. Correct concentration measurement is fundamental in achieving desired chemical properties and behavior in solutions.