Question

Calculate the molarity of each of the following solutions: a. \(0.500\) mole of glucose in \(0.200 \mathrm{~L}\) of solution b. \(36.5 \mathrm{~g}\) of \(\mathrm{HCl}\) in \(1.00 \mathrm{~L}\) of solution c. \(30.0 \mathrm{~g}\) of \(\mathrm{NaOH}\) in \(350 . \mathrm{mL}\) of solution

Short answer

a) 2.50 M; b) 1.00 M; c) 2.14 M.

Step-by-step solution

Step 1a: Identify the formula for molarity

Molarity (M) is given by the formula: \[ M = \frac{n}{V} \]where \( n \) is the number of moles of the solute and \( V \) is the volume of the solution in liters.

Step 2a: Substitute the values for glucose solution

For part (a), the number of moles \( n = 0.500 \) moles, and the volume \( V = 0.200 \) L. Substitute these values into the molarity formula:\[ M_{\text{glucose}} = \frac{0.500 \text{ moles}}{0.200 \text{ L}} = 2.50 \text{ M} \]

Step 3a: Solution for part (a)

The molarity of the glucose solution is \( 2.50 \text{ M} \).

Step 1b: Convert mass to moles

For part (b), first convert the mass of \( \text{HCl} \) to moles. Use the molar mass of \( \text{HCl} \) which is \( 36.5 \text{ g/mol} \):\[ n_{\text{HCl}} = \frac{36.5 \text{ g}}{36.5 \text{ g/mol}} = 1.00 \text{ mole} \]

Step 2b: Substitute the values for HCl solution

Now, use the molarity formula with \( n = 1.00 \) mole and \( V = 1.00 \) L:\[ M_{\text{HCl}} = \frac{1.00 \text{ moles}}{1.00 \text{ L}} = 1.00 \text{ M} \]

Step 3b: Solution for part (b)

The molarity of the HCl solution is \( 1.00 \text{ M} \).

Step 1c: Convert mass to moles

For part (c), first convert the mass of \( \text{NaOH} \) to moles. Use the molar mass of \( \text{NaOH} \) which is \( 40.0 \text{ g/mol} \):\[ n_{\text{NaOH}} = \frac{30.0 \text{ g}}{40.0 \text{ g/mol}} = 0.750 \text{ moles} \]

Step 2c: Convert volume to liters

Convert the volume from mL to L for consistency. Since 1 L = 1000 mL, we have:\[ V = 350 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.350 \text{ L} \]

Step 3c: Substitute the values for NaOH solution

Now, use the molarity formula with \( n = 0.750 \) moles and \( V = 0.350 \) L:\[ M_{\text{NaOH}} = \frac{0.750 \text{ moles}}{0.350 \text{ L}} \~ 2.14 \text{ M} \]

Step 4c: Solution for part (c)

The molarity of the NaOH solution is approximately \( 2.14 \text{ M} \).

Key concepts

Molarity

Molarity is a key concept in chemistry that helps us understand how concentrated a solution is. 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 of the solute, and \( V \) is the volume of the solution in liters.
For example, if we need to calculate the molarity of a glucose solution that has 0.500 moles of glucose in 0.200 L of solution, we substitute the values into the formula to get: \[ M_{\text{glucose}} = \frac{0.500 \text{ moles}}{0.200 \text{ L}} = 2.50 \text{ M} \] This shows that the solution is 2.50 M in glucose.

Mole Conversion

Before we can use the molarity formula, we sometimes need to convert the amount of solute from grams to moles. This is done using the molar mass of the substance, which is the mass of one mole of that substance. The conversion formula is: \[ n = \frac{m}{M_{\text{molar}}} \] where \( n \) is the number of moles, \( m \) is the mass of the substance in grams, and \( M_{\text{molar}} \) is the molar mass in g/mol.
For example, to find the number of moles in 36.5 g of HCl with a molar mass of 36.5 g/mol, we would use: \[ n_{\text{HCl}} = \frac{36.5 \text{ g}}{36.5 \text{ g/mol}} = 1.00 \text{ mole} \] Once we have the number of moles, we can use it in the molarity formula to find the concentration of the solution.

Solution Volume

The volume of the solution must be in liters when calculating molarity. If the volume is given in milliliters, we need to convert it to liters. The conversion factor is: \[ 1 \text{ L} = 1000 \text{ mL} \] To convert milliliters to liters, divide the volume in milliliters by 1000.
For example, if we have 350 mL of solution, we convert it to liters like this: \[ V = 350 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.350 \text{ L} \] This ensures that the volume is correctly represented in the molarity formula. By using consistent units, we can accurately calculate the molarity of the solution.