Vapour pressure of 0.5 molal solution of non-volatile solute in organic solvent at 30°C would be : (Given : vapour pressure of pure organic solvent is 100 torr; the molecular weight of organic solvent is 100 g mol-1).
90.25 torr
Raoult's law for a non-volatile solute gives \(P = x_{solvent}\,P^{\circ}\), so the calculation is really about converting molality to mole fraction.
Step 1 — moles of solute. 0.5 molal means 0.5 mol of solute per 1 kg of solvent.
Step 2 — moles of solvent. With \(M = 100\) g mol-1, one kilogram contains \(\frac{1000}{100} = 10\) mol.
Step 3 — mole fraction of solvent. \(x_{solvent} = \frac{10}{10 + 0.5} = \frac{10}{10.5} = 0.952\).
Step 4 — apply Raoult's law. \(P = 0.952 \times 100 = 95.2\) torr.
Two sanity checks confirm the method. A non-volatile solute can only lower the vapour pressure, so any answer above 100 torr is impossible on physical grounds — which excludes both 125.50 and 119.95 torr immediately. And the solution is dilute, so the lowering should be modest, of the order of 5 torr.
The correctly computed value of 95.2 torr does not appear among the four options; 90.25 torr is the nearest, and 100.25 torr would require the vapour pressure to rise.
Note: the SPPU final answer key marked this question with code 9 ("No option is correct or the question is wrong"), consistent with the calculation above. The answer recorded here is the closest available option and is flagged as key-disputed.
Similar Questions
The property measured in derivative thermogravimetric analysis (DTG) is :
Analysis of four different samples of an alloy yielded 12.12, 12.14, 12.10 and 12.22% of metal. The mean deviation of the result is :
For the first order consecutive reaction :
Which of the following statement is incorrect ?
A given reaction is fitted into the following Arrhenius form :
K2 = 6.0 × 1014 (S-1) . exp \(\left[-\frac{104.4\ (\mathrm{kJ\,mol^{-1}})}{RT}\right]\)
Value of the rate constant at very high temperature would be :
E° value for various ions are as follows :
E°(MnO4-) = +1.51 V
E°(Ag/Ag+) = +0.7996 V
E°(Au/Au+) = +1.692 V
E°(Zn/Zn2+) = -0.761 V
Based on this data, permaganate can be used to oxidize :
The following half cell represents, normal hydrogen electrode (NHE) :
If, we discharge the electrochemical cell rapidly, then :
Consider the reaction :
H2S(g) + \(\tfrac{3}{2}\) O2(g) → SO2(g) + H2O(g), ΔH° = 518.62 kJ
What is the effect of increase of temperature and pressure on the reaction ?
The number of microstates of distributing five quanta of energy among four distinguishable particles is :
The variation of Cp with pressure at constant temperature is given by :
Important Questions from Thermodynamics
- Second law of thermodynamics states that
A. Energy flows from higher concentration to lower concentration.
B. Order becomes disorder during energy transformations.
C. The quality of energy degrades as it is transformed.
D. Degraded energy is entropy, dessipated as waste products or heat.
E. Enthalpy is wasted in energy transformation.
Choose the most appropriate answer from the options given below: Choose the Incorrect Statement:
1 For a system at equilibrium, chemical potential is same in all phases.
2 Gibb's free energy change is zero at equilibrium state of a system.
3 Gibb's free energy change is positive for reaction to occur spontaneously.
4 $\Delta G$ will become more negative with increasing temperature.- In a closed thermodynamic system, across its boundaries
Melting Point of pure ice is-
Which one among the following statements with reference to the properties of water is not correct?
