Question

By dissolving \(5 \mathrm{~g}\) substance in \(50 \mathrm{~g}\) of water, the decrease in freezing point is \(1.2^{\circ} \mathrm{C}\). The molal depression constant is \(1.85^{\circ} \mathrm{kg} \mathrm{mol}^{-1} .\) The molecular weight of substance is (a) \(105.4\) (b) \(118.2\) (c) \(137.2\) (d) \(154.2\)

Short answer

The molecular weight of the substance is \(154.2\, \text{g/mol}\).

Step-by-step solution

Understanding the Problem

We are given the mass of the solute and solvent, the change in the freezing point, and the molal depression constant. We need to find the molecular weight (molar mass) of the unknown substance.

Calculate Molality

Molality (m) is calculated using the formula \( m = \frac{\Delta T_f}{K_f} \), where \( \Delta T_f \) is the freezing point depression, and \( K_f \) is the molal depression constant. Substituting the given values, we have:\[ m = \frac{1.2}{1.85} = 0.64865 \, \text{mol/kg} \]

Determine Moles of Solute

The number of moles of solute can be determined using the relation:\[ \text{moles of solute} = m \times \text{mass of solvent in kg} \]Given that the mass of the water is \(50\, \text{g} = 0.050\, \text{kg} \), the moles of solute is: \[ 0.64865 \, \text{mol/kg} \times 0.050 \, \text{kg} = 0.0324325 \, \text{mol} \]

Calculate Molar Mass

Using the number of moles from the previous step, the molar mass is calculated by the equation: \[ \text{Molar Mass} = \frac{\text{mass of solute}}{\text{moles of solute}} \]Substituting the values:\[ \text{Molar Mass} = \frac{5 \, \text{g}}{0.0324325 \, \text{mol}} = 154.2 \, \text{g/mol} \]

Verify Answer

Review the calculations to ensure all values and units are correctly aligned with the problem statement. The molar mass calculated matches one of the given options: (d) 154.2.

Key concepts

Molecular Weight Calculation

When it comes to determining the molecular weight of a substance, the concept revolves around calculating how much the substance weighs when we have one mole of it. In this exercise, we start by identifying the amount of solute in grams and the number of moles present. The molecular weight ( M ) is given by the equation:
\[ M = \frac{\text{mass of solute}}{\text{moles of solute}} \]
Knowing the mass of the solute, which is given as 5 grams, and moles that were earlier calculated as 0.0324325 mol, you can substitute these values into the equation to find the molecular weight.

• This results in \( M = \frac{5 \, \text{g}}{0.0324325 \, \text{mol}} \), leading to a molecular weight of 154.2 g/mol

Understanding the molecular weight helps in identifying unknown compounds and is fundamental in stoichiometry, the branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions.

Molality

Molality is a key concept when dealing with solutions and calculating freezing point depression. It measures how many moles of solute are present per kilogram of solvent. Unlike molarity, which depends on the volume of the solution, molality is based on the mass allowing it to remain unaffected by changes in temperature and pressure. The formula for molality ( m ) is:
\[ m = \frac{\Delta T_f}{K_f} \]
Where \( \Delta T_f \) is the freezing point depression, and \( K_f \) is the molal depression constant. In the given exercise:

• \( \Delta T_f = 1.2^{\circ} \text{C} \)
• \( K_f = 1.85^{\circ} \text{kg/mol} \)
• This results in a molality of \( m = \frac{1.2}{1.85} = 0.64865 \, \text{mol/kg} \)

Molality becomes particularly useful in studying colligative properties as it remains unaffected by any changes in external conditions.

Freezing Point Depression Constant

The freezing point depression constant ( K_f ) is an essential factor in calculating how the freezing point of a liquid solvent decreases when a solute is dissolved into it. It reflects the change in freezing point per molal concentration of the solution. Here’s how it works:

• The constant \( K_f \) can vary depending on the solvent. Common solvents in chemistry like water or benzene have specific depression constants that are known through experiments.
• It's necessary in calculating the freezing point depression, which is a colligative property. This means it depends on the number of particles in a solution, not on their identity.
• Freezing point depression calculations are useful in many approaches, like identifying substances or calculating molecular weights, much like in our example problem.

Having a grip on the freezing point depression constant allows chemists to predict and measure the effects of solutes on different solvents effectively.