Problem 95 - A Complete Resource Book in Chemistry for JEE Main

Question

120 gram of urea is present in 5 litre of solution. The active mass of urea is (a) \(0.06\) (b) \(0.2\) (c) \(0.4\) (d) \(1.4\)

Step-by-Step Solution

Answer

The active mass of urea is 0.4 M, option (c).

Step 1: Identify the Given Data

First, identify the information given in the problem. We know that there is 120 grams of urea dissolved in a 5-liter solution. We need to calculate the active mass, which is the molarity of the solution.

Step 2: Calculate Molar Mass of Urea

Urea ( ext{CO}( ext{NH}_2)_2) has a molar mass of: 1 Carbon (C) = 12 g/mol, 1 Oxygen (O) = 16 g/mol, 2 Nitrogen (N) = 28 g/mol (2 x 14 g/mol), and 4 Hydrogen (H) = 4 g/mol (4 x 1 g/mol). Add these together: 12 + 16 + 28 + 4 = 60 g/mol.

Step 3: Calculate Moles of Urea

To find the number of moles of urea, divide the mass by the molar mass: \( \text{Moles of urea} = \frac{120}{60} = 2 \text{ moles} \).

Step 4: Calculate the Active Mass (Molarity)

Molarity is calculated by dividing the number of moles of solute by the volume of the solution in liters: \( \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{2}{5} = 0.4 \text{ M} \).

Key Concepts

Molar mass calculation

Calculating the molar mass of a compound is the first step in converting grams to moles, which is essential for determining solution concentration. In this example, we are dealing with urea, a common compound with the chemical formula \( \text{CO}(\text{NH}_2)_2 \).

To find the molar mass, you add up the atomic masses of all atoms in the compound:

• Carbon (C): 1 atom \( \times 12 \text{ g/mol} = 12 \text{ g/mol} \)
• Oxygen (O): 1 atom \( \times 16 \text{ g/mol} = 16 \text{ g/mol} \)
• Nitrogen (N): 2 atoms \( \times 14 \text{ g/mol} = 28 \text{ g/mol} \)
• Hydrogen (H): 4 atoms \( \times 1 \text{ g/mol} = 4 \text{ g/mol} \)

Add these values together 12 + 16 + 28 + 4 = 60 \text{ g/mol}. Here, 60 g/mol is the molar mass of urea.

Knowing the correct molar mass enables us to further calculate the number of moles present in a given mass of the compound.

Moles of solute

To determine molarity, it's crucial to know how many moles of solute are present. Moles are a way of measuring the amount of a substance, much like counting dozens. The formula to find the number of moles is:

\[ \text{Moles of solute} = \frac{\text{mass of solute in grams}}{\text{molar mass of the solute in g/mol}} \]
Applying this to our example with urea, we take the mass of urea, which is 120 grams, and divide it by its molar mass, 60 g/mol:

\[ \text{Moles of urea} = \frac{120}{60} = 2 \text{ moles} \]
Understanding this step is important because the number of moles directly influences the molarity and concentration calculations that follow.

Volume of solution in liters

Volume plays a critical role in calculating molarity, as molarity is defined as the number of moles of solute per liter of solution. When working with solutions, it's essential to understand that volume should always be expressed in liters to correctly calculate molarity.

In our scenario, the solution volume is given as 5 liters. This aligns with the calculation for molarity:

\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
Substituting the known values:

\[ \text{Molarity} = \frac{2 \text{ moles}}{5 \text{ liters}} = 0.4 \text{ M} \]
Therefore, the molarity, or active mass, of the urea solution is 0.4 M. It’s always important to ensure that the solution volume is correctly converted to liters to avoid errors in molarity calculations.