Problem 15 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Dissolving \(120 \mathrm{~g}\) of urea (mol. wt. 60 ) in \(1000 \mathrm{~g}\) of water gave a solution of density \(1.15 \mathrm{~g} / \mathrm{mL}\). The molarity of the solution is (a) \(1.78 \mathrm{M}\) (b) \(2.00 \mathrm{M}\) (c) \(2.05 \mathrm{M}\) (d) \(2.22 \mathrm{M}\)

Step-by-Step Solution

Answer

The molarity of the solution is \(2.05 \text{ M}\).

Step 1: Calculate the Moles of Urea

First, we need to calculate the number of moles of urea. Use the formula for moles: \( \text{moles} = \frac{\text{mass (g)}}{\text{molecular weight (g/mol)}} \). We have \( 120 \, \text{g} \) of urea and a molecular weight of \( 60 \, \text{g/mol} \). Thus, the number of moles of urea is: \[ \text{moles of urea} = \frac{120}{60} = 2 \text{ moles} \].

Step 2: Calculate the Volume of the Solution

Next, we calculate the volume of the solution using its density. The total mass of the solution is the sum of the mass of the solute (urea) and solvent (water). Therefore, the mass of the solution is \( 120 \, \text{g} + 1000 \, \text{g} = 1120 \, \text{g} \). Given the density of the solution is \( 1.15 \, \text{g/mL} \), we calculate the volume by using \( \text{volume} = \frac{\text{mass}}{\text{density}} \). Thus, \[ \text{volume of solution} = \frac{1120}{1.15} \approx 973.91 \, \text{mL} \].

Step 3: Convert Volume to Liters

We need to convert the volume from milliliters to liters, as molarity is expressed in moles per liter. Remember, \( 1 \, \text{L} = 1000 \, \text{mL} \). Therefore: \[ \text{volume in liters} = \frac{973.91}{1000} \approx 0.974 \text{ L} \].

Step 4: Calculate the Molarity of the Solution

Molarity \( M \) is defined as the number of moles of solute per liter of solution. We have \( 2 \) moles of urea and \( 0.974 \text{ L} \) of solution. Therefore, the molarity is: \[ M = \frac{2}{0.974} \approx 2.05 \, \text{M} \].

Key Concepts

Volume Conversion

In chemistry, volume conversion is important, especially when dealing with measurements in different units. Molarity calculation requires the final volume to be in liters. This is necessary because molarity is expressed in moles per liter.

A common conversion is from milliliters (mL) to liters (L). Consider the fact that 1000 mL equals 1 L. To convert a volume from mL to L, you need to divide the mL value by 1000. For example, if you have a volume of 973.91 mL, converting it to liters involves the calculation: \[ \text{volume in liters} = \frac{973.91}{1000} \approx 0.974 \, \text{L} \]
By performing this conversion, volumes are standardized to a common unit, making it easier to calculate molarity.

Density of Solutions

Understanding the density of a solution is crucial for determining its volume when only the mass is known. Density is defined as the mass per unit volume and is usually expressed in grams per milliliter (g/mL).

For example, given a solution with a total mass of 1120 g and a density of 1.15 g/mL, you can find its volume by using the formula: \[ \text{volume} = \frac{\text{mass}}{\text{density}} \]

• Mass of the solution: 1120 g
• Density: 1.15 g/mL
• Volume calculation: \[ \text{volume of solution} = \frac{1120}{1.15} \approx 973.91 \, \text{mL} \]

This calculation shows how density acts as a bridge between mass and volume, allowing for the determination of one when the other two are known.

Moles Calculation

Calculating the number of moles is a fundamental step in chemistry, necessary for determining the concentration of solutions. Moles measure the amount of substance, and are calculated using the mass of the substance and its molecular weight.

The formula for finding moles is: \[ \text{moles} = \frac{\text{mass (g)}}{\text{molecular weight (g/mol)}} \]

• Given mass of urea: 120 g
• Molecular weight of urea: 60 g/mol
• Calculation: \[ \text{moles of urea} = \frac{120}{60} = 2 \text{ moles} \]

This calculation is crucial for establishing how much of a substance is present in a chemical reaction or solution, paving the way for further calculations like molarity.