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0.24 g of a volatile liquid on vaporization gives 45 ml of vapours at STP.What will be the vapour density of the substance ?(Density of H₂ = 0.089 gL⁻1)
  • a)
    95.39
  • b)
    39.95
  • c)
    99.53
  • d)
    59.93
Correct answer is option 'D'. Can you explain this answer?
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0.24 g of a volatile liquid on vaporization gives 45 ml of vapours at ...
Given:
Mass of the volatile liquid = 0.24 g
Volume of vapors at STP = 45 ml = 45 cm^3
Density of hydrogen (H2) gas = 0.089 g/L

To find: Vapour density of the substance

Vapour density is defined as the ratio of the molar mass of a substance to the molar mass of hydrogen gas.

We can calculate the molar mass of the substance using the ideal gas equation:
PV = nRT

Where:
P = Pressure (in atm)
V = Volume (in L)
n = Number of moles of the substance
R = Ideal gas constant (0.082 L.atm/mol.K)
T = Temperature (in K)

Since the volume is given at STP, we can assume the pressure to be 1 atm and the temperature to be 273 K.

Using the ideal gas equation, we can rearrange it to solve for n:
n = PV / RT

Since the volume is given in cm^3, we need to convert it to L:
Volume = 45 cm^3 = 45/1000 L = 0.045 L

Substituting the given values into the equation:
n = (1 atm) (0.045 L) / [(0.082 L.atm/mol.K) (273 K)]
n = 0.00176 mol

The molar mass of the substance can be calculated using the formula:
Molar mass = mass / number of moles

Substituting the given values into the equation:
Molar mass = 0.24 g / 0.00176 mol
Molar mass = 136.36 g/mol

The molar mass of hydrogen gas (H2) is 2 g/mol.

Finally, we can calculate the vapour density of the substance:
Vapour density = Molar mass / Molar mass of H2
Vapour density = 136.36 g/mol / 2 g/mol
Vapour density = 68.18

Since the options provided are in g/L, we need to convert the vapour density to g/L by multiplying it by the density of hydrogen gas:
Vapour density = 68.18 * 0.089 g/L
Vapour density = 6.07 g/L

Therefore, the correct answer is option 'D': 59.93 g/L.
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Direction: Read the passage given below and answer the following questions: Boiling point or freezing point of liquid solution would be affected by the dissolved solids in the liquid phase. A soluble solid in solution has the effect of raising its boiling point and depressing its freezing point. The addition of non-volatile substances to a solvent decreases the vapour pressure and the added solute particles affect the formation of pure solvent crystals. According to many researches the decrease in freezing point directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapour pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity. Q.When a non volatile solid is added to pure water it will

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0.24 g of a volatile liquid on vaporization gives 45 ml of vapours at STP.What will be the vapour density of the substance ?(Density of H = 0.089 gL1)a)95.39b)39.95c)99.53d)59.93Correct answer is option 'D'. Can you explain this answer?
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