63g of a compound (Mol.Wt.=126) was dissolved in 500g of distilled water.The density of the resultant solution is 1.126g/ml. The molarity of the solution is
- 1.25M
- 1.0M
- 0.75M
- 1.1M
The Correct Option is B
Approach Solution - 1
Given:
- Mass of compound = 63 g
- Molecular weight of compound = 126 g/mol
- Mass of water = 500 g
- Density of solution = 1.126 g/ml
First, let's find the volume of the solution. The density formula is \( \text{density} = \frac{\text{mass}}{\text{volume}} \), so the volume of the solution is:
\[ \text{Volume of solution} = \frac{\text{mass of solution}}{\text{density of solution}} = \frac{63 g + 500 g}{1.126 g/ml} \]
\[ \text{Volume of solution} = \frac{563 g}{1.126 g/ml} \]
\[ \text{Volume of solution} = 500 ml \]
Since the density of water is 1 g/ml, the mass of water in the solution is 500 g. This means the mass of the compound in the solution is \( 63 \text{ g} \). Since the compound's molecular weight is 126 g/mol, the number of moles of the compound in the solution is:
\[ \text{Moles of compound} = \frac{\text{Mass of compound}}{\text{Molecular weight of compound}} \]
\[ \text{Moles of compound} = \frac{63 \text{ g}}{126 \text{ g/mol}} \]
\[ \text{Moles of compound} = 0.5 \text{ mol} \]
Now, the molarity of the solution is given by the formula:
\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]
Since we have 0.5 moles of compound and the volume of the solution is 500 ml (0.5 liters), the molarity is:
\[ \text{Molarity} = \frac{0.5 \text{ mol}}{0.5 \text{ L}} \]
\[ \text{Molarity} = 1.0 \text{ M} \]
Therefore, the molarity of the solution is 1.0 M.
The correct answer is option (B): 1.0M
Approach Solution -2
Given:
- Mass of solute = 63 g
- Molar mass = 126 g/mol
- Mass of water = 500 g
- Density of solution = 1.126 g/mL
Step 1: Calculate moles of solute
\[ \text{Moles} = \frac{63}{126} = 0.5\ \text{mol} \]
Step 2: Calculate total mass of solution
\[ \text{Total mass} = 63\ \text{g (solute)} + 500\ \text{g (solvent)} = 563\ \text{g} \]
Volume of solution:
\[ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{563}{1.126} \approx 500\ \text{mL} = 0.5\ \text{L} \]
Step 3: Calculate molarity
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{0.5}{0.5} = 1.0\ \text{M} \]
Final Answer: 1.0 M
Top Questions on Solutions
- Calculate the boiling point of a solution containing 0.61 g of benzoic acid (122 \( g mol^{-1} \)) in 5 g of \( CS_2 \) in which it dimerises to 88%. \( T_b^\circ(CS_2) = 46.2^\circ C, K_b = 2.3 K kg mol^{-1} \).
- State Raoult’s Law for volatile solutes and explain its deviations.
- Define Colligative Properties and provide two examples.
- What type of deviation from Raoult’s law is shown by mixture of ethanol and acetone? Give reason. What will happen to the boiling point of the solution on mixing ethanol and acetone?
- Assertion (A) : Components of azeotropes are easily separated by fractional distillation.
Reason (R) : Components of an azeotrope have same composition in liquid and vapour phase.
Questions Asked in WBJEE exam
What are the charges stored in the \( 1\,\mu\text{F} \) and \( 2\,\mu\text{F} \) capacitors in the circuit once current becomes steady?
- WBJEE - 2026
- Capacitors and Capacitance
Which one among the following compounds will most readily be dehydrated under acidic condition?
- WBJEE - 2026
- Organic Reactions
Manufacturers supply a zener diode with zener voltage \( V_z=5.6\,\text{V} \) and maximum power dissipation \( P_{\max}=\frac14\,\text{W} \). This zener diode is used in the circuit shown. Calculate the minimum value of the resistance \( R_s \) so that the zener diode will not burn when the input voltage is \( V_{in}=10\,\text{V} \).
- WBJEE - 2026
- Zener Diodes
Two charges \( +q \) and \( -q \) are placed at points \( A \) and \( B \) respectively which are at a distance \( 2L \) apart. \( C \) is the midpoint of \( AB \). The work done in moving a charge \( +Q \) along the semicircle CSD (\( W_1 \)) and along the line CBD (\( W_2 \)) are
- WBJEE - 2026
- electrostatic potential and capacitance
A piece of granite floats at the interface of mercury and water. If the densities of granite, water and mercury are \( \rho, \rho_1, \rho_2 \) respectively, the ratio of volume of granite in water to that in mercury is
- WBJEE - 2026
- Buoyancy and Floatation
Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.