Question:
The relative lowering of vapor pressure of an aqueous solution containing a non-volatile solute is 0.0225. The mole fraction of the non-volatile solute is:
The relative lowering of vapor pressure of an aqueous solution containing a non-volatile solute is 0.0225. The mole fraction of the non-volatile solute is:
Show Hint
For dilute solutions, the mole fraction of solute can be approximated as the relative lowering of vapor pressure.
Updated On: Feb 25, 2025
- 0.80
- 0.725
- 0.15
- 0.0225
Hide Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Relative lowering of vapor pressure is given by Raoult’s Law:
\[
\frac{P_0 - P}{P_0} = X_B
\]
where:
- \( P_0 \) is the vapor pressure of pure solvent,
- \( P \) is the vapor pressure of the solution,
- \( X_B \) is the mole fraction of the solute.
Since relative lowering of vapor pressure is given as 0.0225, we directly get:
\[
X_B = 0.0225
\]
Was this answer helpful?
0
0
Top Questions on Solutions
- Shweta mixed two liquids A and B of 10 mL each. After mixing, the volume of the solution was found to be 20.2 mL.
(i) Why was there a volume change after mixing the liquids?
(ii) Will there be an increase or decrease of temperature after mixing?
(iii) Give one example for this type of solution. - Give reasons:
(a) Cooking is faster in a pressure cooker than in an open pan.
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y? - In case of association, abnormal molar mass of solute will:
- An aqueous solution of 0.1 M HA shows depression in freezing point of 0.2°C. If $ K_f $ (H$_2$O) = 1.86 K kg mol$^{-1}$ and assuming molarity = molality, find the dissociation constant of HA.
- If equal volumes of AB and XY (both are salts) aqueous solutions are mixed, which of the following combination will give precipitate of AY, at 300 K?
View More Questions
Questions Asked in CBSE CLASS XII exam
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
- CBSE CLASS XII - 2025
- Integration
- A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).
- CBSE CLASS XII - 2025
- Magnetic moment
- A system of two conductors is placed in air and they have net charge of \( +80 \, \mu C \) and \( -80 \, \mu C \) which causes a potential difference of 16 V between them.
(1) Find the capacitance of the system.
(2) If the air between the capacitor is replaced by a dielectric medium of dielectric constant 3, what will be the potential difference between the two conductors?
(3) If the charges on two conductors are changed to +160µC and −160µC, will the capacitance of the system change? Give reason for your answer.- CBSE CLASS XII - 2025
- Capacitors and Capacitance
- Write the mathematical forms of three postulates of Bohr’s theory of the hydrogen atom. Using them prove that, for an electron revolving in the \( n \)-th orbit,
(a) the radius of the orbit is proportional to \( n^2 \), and
(b) the total energy of the atom is proportional to \( \frac{1}{n^2} \).- CBSE CLASS XII - 2025
- Atomic Structure
- VX Ltd. issued 30,000, 8% debentures of ₹ 100 each at a discount of 10% redeemable at a certain rate of premium. On issue of these debentures, ‘Loss on issue of debentures account’ was debited with ₹ 4,50,000. The amount of premium on redemption of debentures was___.
- CBSE CLASS XII - 2025
- Debentures
View More Questions