Question

How much (a) glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in grams, must be dissolved in water to produce \(75.0 \mathrm{mL}\) of \(0.350 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} ?\) (b) methanol, \(\mathrm{CH}_{3} \mathrm{OH} \quad(d=0.792 \mathrm{g} / \mathrm{mL}),\) in milli- liters, must be dissolved in water to produce 2.25 L of \(0.485 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH} ?\)

Short answer

Thus, (a) 4.73 grams of glucose and (b) 44.1 milliliters of methanol are needed to prepare the solutions.

Step-by-step solution

Find amount of glucose in moles

To find the amount of glucose in moles, we can use the molarity equation: M = n / V. Where M is the molarity, n is the number of moles, and V is the volume. We are given M = 0.350 M and V = 75.0 mL or 0.075 L (converted from mL to L). Let's solve for n: n = M * V = 0.350 moles/L * 0.075 L = 0.02625 moles.

Convert moles to grams

To convert the number of moles to mass in grams, we can use the molecular weight of glucose which is 180.16 g/mol. The mass of glucose = number of moles * molecular weight. So the mass of glucose = 0.02625 mol * 180.16 g/mol = 4.73 g.

Find amount of methanol in moles

Similarly, for methanol, we use the molarity equation and given values: M = 0.485 M and V = 2.25 L. So, the amount of methanol in moles = M * V = 0.485 moles/L * 2.25 L = 1.09 moles.

Find volume of methanol

To find the volume of methanol, we use its density: d = m / V. We are given d = 0.792 g/mL. Rearranging the formula to solve for V: V = m / d. But first we need to calculate m: the mass of methanol = number of moles * molecular weight = 1.09 mol * 32.04 g/mol = 34.92 g. Now, we can find V: V = m / d = 34.92 g / 0.792 g/mL = 44.1 mL

Key concepts

Molarity

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. The formula for molarity is: \[ M = \frac{n}{V} \]where \( M \) is the molarity, \( n \) is the number of moles, and \( V \) is the volume of the solution in liters. Knowing this helps in predicting how substances will react when mixed. In the exercise, glucose and methanol molarity values were used to determine the amount needed for specific solution volumes. This practice is common in making laboratory solutions and mixing chemicals.

Key points to remember:

• Molarity units are moles per liter (mol/L).
• It helps calculate how much solute is needed for a desired concentration.
• Conversions between mL and L are often required when working with solutions.

Molecular Weight

Molecular weight is the mass of a single molecule of a substance, expressed in atomic mass units (amu). It is crucial when you want to convert between moles and grams because it acts as a bridge between the two. The molecular weight of glucose, for instance, is 180.16 g/mol.

This value is useful because:

• It allows you to convert moles to grams and vice versa using the formula:

\[ \text{Mass (g)} = \text{Moles} \times \text{Molecular Weight (g/mol)} \]
This calculation is vital in determining the exact weight of a compound needed in a solution, as seen in calculating the amount of glucose or methanol needed in the exercise.

Density

Density is a property that relates a substance's mass to its volume, commonly expressed as grams per milliliter (g/mL). It enables the conversion of mass into volume and vice versa, which is essential when you need a specific amount of liquid chemical, as was done with methanol.

The formula for density is \[ d = \frac{m}{V} \]where \( d \) is the density, \( m \) is the mass, and \( V \) is the volume. Rearranging the formula can help find the volume if the mass and density are known:

• To find the mass, multiply volume by density.
• To find volume, divide mass by density.

In calculations with methanol, the density was crucial in determining how many milliliters were needed.

Volume Conversion

Volume conversion is essential when working with solutions because laboratory equipment often measures volume in milliliters (mL), but calculations are generally performed using liters (L). To convert between these units, remember:

• 1 L = 1000 mL
• Simply divide by 1000 to convert mL to L, or multiply by 1000 to convert L to mL.

This conversion ensures that you use consistent units across your calculations, which is crucial in accurately determining solute concentrations, as was seen in converting 75.0 mL to 0.075 L for the glucose solution.