Question

A solution contains 20.4g glucose (C6H12O6) dissolved in 0.640-L of water. What is the molality of the solution (Assume a density of 1.00 g/mL for water)?

Answer 1

The molality of the solution is 0.600 mol/kg.

Step-by-step explanation

To determine the molality of the solution, we can use the formula:

`\[ \text{Molality} (m) = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \]`

First, we need to find the moles of glucose (C₆H₁₂O₆). The molar mass of glucose is calculated as follows:

`\[ \text{Molar mass of C₆H₁₂O₆} = (6 * \text{atomic mass of C}) + (12 * \text{atomic mass of H}) + (6 * \text{atomic mass of O}) \]`

`\[ = (6 * 12.01) + (12 * 1.01) + (6 * 16.00) \]`

`\[ = 72.06 + 12.12 + 96.00 \]`

`\[ = 180.18 \, \text{g/mol} \]`

Now, we can find the moles of glucose using the given mass:

`\[ \text{moles of C₆H₁₂O₆} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} \]`

`\[ = \frac{20.4 \, \text{g}}{180.18 \, \text{g/mol}} \]`

`\[ \approx 0.113 \, \text{mol} \]`

Next, we find the mass of the solvent (water) using its volume and density:

`\[ \text{mass of water} = \text{volume of water} * \text{density of water} \]`

`\[ = 0.640 \, \text{L} * 1.00 \, \text{g/mL} \]`

`\[ = 0.640 \, \text{kg} \]`

Now, we can calculate molality:

`\[ \text{Molality} = \frac{0.113 \, \text{mol}}{0.640 \, \text{kg}} \]`

`\[ \approx 0.600 \, \text{mol/kg} \]`

So, the molality of the solution is 0.600 mol/kg.