Question:
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
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
Updated On: Jun 24, 2024
- 1.25M
- 1.0M
- 0.75M
- 1.1M
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
To find the molarity of the solution, we first need to calculate the number of moles of the compound and then use the formula for molarity.
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
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
Was this answer helpful?
5
6
Top Questions on Solutions
- Arrange the following in increasing order of solubility product: \[ {Ca(OH)}_2, {AgBr}, {PbS}, {HgS} \]
- Concentrated nitric acid is labelled as 75%by mass. The volume in mL of the solution which contains 30 g of nitric acid is: Given: Density of nitric acid solution is 1.25 g/mL.
- Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is:
- 20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is \(\_\_\_\_\_\) x \(10^{-2}\)M. (Nearest integer)
- 0.1 M solution of KI reacts with excess of \( \text{H}_2\text{SO}_4 \) and KIO\(_3\), according to the equation: \[ 5I^- + 6H^+ \rightarrow 3I_2 + 3H_2O \] Identify the correct statements:
View More Questions
Questions Asked in WBJEE exam
- Two straight conducting plates form an angle θ where their ends are joined. A conducting bar in contact with the plates and forming an isosceles triangle with them, starts at the vertex at time t = 0 and moves with constant velocity v to the right as shown in the figure. A magnetic field B points out of the page. The magnitude of emf induced at t = 1 second will be:
- WBJEE - 2024
- Electromagnetic induction
- A 2V cell is connected across the points A and B as shown in the figure. Assume that the resistance of each diode is zero in forward bias and infinity in reverse bias. The current supplied by the cell is:
- WBJEE - 2024
- Current electricity
- Consider a circuit where a cell of emf \( E_0 \) and internal resistance \( r \) is connected across the terminal A and B as shown in the figure. The value of \( R \) for which the power generated in the circuit is maximum, is given by:
- WBJEE - 2024
- Current electricity
- In a series LCR circuit, the rms voltage across the resistor and the capacitor are \(30 \, \text{V}\) and \(90 \, \text{V}\) respectively. If the applied voltage is \(50\sqrt{2}\sin\omega t\), then the peak voltage across the inductor is:
- WBJEE - 2024
- Alternating current
- Which of the following statement(s) is/are true in respect of nuclear binding energy?
(i) The mass energy of a nucleus is larger than the total mass energy of its individual protons and neutrons.
(ii) If a nucleus could be separated into its nucleons, an energy equal to the binding energy would have to be transferred to the particles during the separating process.
(iii) The binding energy is a measure of how well the nucleons in a nucleus are held together.
(iv) The nuclear fission is somehow related to acquiring higher binding energy.- WBJEE - 2024
- Nuclear physics
View More Questions
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.