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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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