Question

By how much will 50 grams of water have its freezing point depressed if you add 30 grams (molecular weight 80 ) of glucose to it?

Short answer

The freezing point of the water is depressed by 13.95 °C when 30 grams of glucose is added to it.

Step-by-step solution

Calculate the moles of glucose

First, we need to determine the moles of glucose in the sample. The number of moles can be found using the formula: moles = mass / molecular_weight Given mass of glucose = 30 grams Molecular weight of glucose = 80 moles(glucose) = \(\frac{30}{80}\) = 0.375 moles

Calculate molality (m)

Molality is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is glucose and the solvent is water. Molality (m) = moles(glucose) / mass(water in kg) The mass of water is 50 grams, which is equal to 0.05 kg. So, the molality of the glucose solution is: m = \(\frac{0.375 \text{ moles}}{0.05 \text{ kg}}\) = 7.5 mol/kg

Use the freezing point depression formula

Now we can use the freezing point depression formula: ΔT_f = K_f * m The cryoscopic constant, K_f, for water is 1.86 °C/mol/kg. So, we can plug this value along with the calculated molality (m = 7.5 mol/kg) into the formula: ΔT_f = 1.86 °C/mol/kg * 7.5 mol/kg

Calculate the final freezing point depression

Now we just need to calculate the product to get the final freezing point depression: ΔT_f = 1.86 °C/mol/kg * 7.5 mol/kg = 13.95 °C Thus, the freezing point of the water is depressed by 13.95 °C when 30 grams of glucose is added to it.

Key concepts

Moles of Solute

Understanding the concept of moles is crucial in chemistry. When we talk about 'moles of solute,' we're referring to the amount of a substance. This quantity is expressed in "moles," a fundamental unit in chemistry. To find the moles of a substance, we use the formula \( \text{moles} = \frac{\text{mass}}{\text{molecular weight}} \).
In our example, we had 30 grams of glucose, with a molecular weight of 80. By plugging these numbers into our formula, we calculated 0.375 moles of glucose. - **Why is this Important?** Calculating moles helps us determine how much impact a solute will have when dissolved in a solvent, like how glucose affects water's freezing point in our example. - **Interpreting Moles in Context:** Think of moles as counting how many molecules you have, similar to how you'd count dozens for eggs. More moles mean more particles affecting the solution's properties.

Molality

Molality is a concentration unit that's slightly different from more familiar units like molarity. It measures how many moles of solute exist per kilogram of solvent. The formula to calculate molality is:\[\text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}\]- **Why Molality?** Molality is useful to use because it doesn't change with temperature. This makes it perfect for involved calculations, like those concerning freezing point depression.In our exercise, 0.375 moles of glucose is dissolved in 50 grams (or 0.05 kg) of water. Using the formula above, we calculated a molality of 7.5 mol/kg. This calculation is essential for finding out how the presence of the solute affects the solvent's properties, particularly its freezing point.- **Visualizing Molality:** Imagine pouring a set number of sugar packets into a set amount of water. Molality describes how packed those packets are in that water, without worrying about whether it’s hot or cold.

Cryoscopic Constant

The cryoscopic constant \(K_f\) is a property of the solvent used in calculating the freezing point depression of solutions. Each solvent has its own cryoscopic constant, which shows how much the freezing point of the solvent will lower per molal concentration of solution.- **Why Use the Cryoscopic Constant?** It's vital for predicting and calculating how much the freezing point will decrease when a solute is added to a solvent. - **In Our Example:** For water, the cryoscopic constant used is 1.86 °C/mol/kg. We applied this in the freezing point depression equation:\[\Delta T_f = K_f \times m\]In the calculation, the constant '1.86' is multiplied by the molality '7.5 mol/kg' of our solution, showing why the solution's freezing point changes. This calculation gives us a clear picture of how much cold weather effects can involve solute addition and how we can predict the resulting temperature shift. This understanding helps in predicting reactions in chemical industries and even food preservation methodologies.