At 50 °C, the vapour pressure of pure benzene is 268 torr. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of 167 torr at the same temperature is
At 50 °C, the vapour pressure of pure benzene is 268 torr. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of 167 torr at the same temperature is
0.505
0.705
0.605
0.405
The Correct Option is C
Solution and Explanation
The correct option is: (C): 0.605.
The decrease in vapor pressure when a non-volatile solute is added to a solvent follows Raoult's law. According to Raoult's law, the vapor pressure of a solvent in a solution is proportional to its mole fraction in the solution.
The formula for Raoult's law is:
P_solvent = x_solvent * P°_solvent
Where:
- P_solvent is the vapor pressure of the solvent in the solution
- x_solvent is the mole fraction of the solvent in the solution
- P°_solvent is the vapor pressure of the pure solvent
In this case, you want to find the mole fraction of benzene (the solvent) that results in a vapor pressure of 167 torr at 50 °C. You're given the vapor pressure of pure benzene as 268 torr at the same temperature.
Rearranging the Raoult's law equation to solve for x_solvent:
x_solvent = P_solvent / P°_solvent
Plug in the values:
x_solvent = 167 torr / 268 torr ≈ 0.6231
Now, the mole fraction of the solute (non-volatile solute) can be calculated as:
x_solute = 1 - x_solvent ≈ 1 - 0.6231 ≈ 0.3769
The number of moles of non-volatile solute per mole of benzene can be calculated using the mole fractions:
Moles of solute / Moles of benzene = x_solute / x_solvent
Moles of solute / 1 = 0.3769 / 0.6231
Moles of solute ≈ 0.605
So, the answer is approximately 0.605
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:
Questions Asked in TS EAMCET exam
- The number of ways in which 15 identical gold coins can be distributed among 3 persons such that each one gets at least 3 gold coins is
- TS EAMCET - 2024
- permutations and combinations
- Identify the incorrect pair:
- TS EAMCET - 2024
- Plant Physiology
- The internal and external diameters of a hollow cylinder measured with vernier calipers are (5.73 \(\pm\) 0.01) cm and (6.01 \(\pm\) 0.01) cm respectively. Then the thickness of the cylinder wall is
- TS EAMCET - 2024
- Units and measurement
- f is a real valued function satisfying the relation \(f\bigl(3x + \tfrac{1}{2x}\bigr) \;=\; 9x^2 + \tfrac{1}{4x^2}\). If \(f\bigl(x + \tfrac{1}{x}\bigr) = 1\) then \(x =\) ?
- TS EAMCET - 2024
- Relations and functions
- A body is projected from the ground at an angle of \( \tan^{-1}\sqrt{7} \) with the horizontal. At half of the maximum height, the speed of the body is \( n \) times the speed of projection. The value of \( n \) is
- TS EAMCET - 2024
- laws of motion
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.