Question

200 cm3 of which of following fluorocarbons will weigh 0.5 gm at 400 K and 0.821 atmospheric
pressure?
(2) C3F6
(3) C2F4
(1) C2F6
(4) C3F4
​

Answer 1

The correct option for the fluorocarbons that weigh 0.5 gm at 400 K and 0.821 atmospheric pressure is (3) C2F4.

Given:

200 cm3 of fluorocarbons weigh 0.5 gm at 400 K and 0.821 atmospheric.

To Find:

The fluorocarbons that weigh 0.5 gm 0 K and 0.821 atmospheric pressure.

Solution:

To find the fluorocarbons that weigh 0.5 gm at 400 K and 0.821 atmospheric pressure is (3) C2F4.

As we know,

According to the ideal gas equation:

$pv = nrt$

$pv = \frac{w}{m} rt$

$m = \frac{wrt}{pv}$

Here, m is the molecular mass, p is the pressure in the atmosphere and w is the given mass.

v is the volume in liters

t is the temperature in kelvin

r is the universal gas constant and its value is 0.0821 atmL/K

Now,

According to the question:

Pressure = 0.821 atm

Volume = 200 cm³ or 0.2 L

T = 400 k

w = 0.5g

(Because 1 liter = 1000cm³)

Now,

Putting values of p, v, T, r, and w in the above formula we get,

Molecular mass (m) =

$m = \frac{0.5 \times 0.0821 \times 400}{0.821 \times0.2 } = \frac{0.5 \times 400}{2} = \frac{200}{2} = 100g$

Now,

The molecular mass of C2F4 is 100 g because

C2F4 = 2 (12) + 4(19) = 24 + 76 = 100g

Henceforth, the correct option for the fluorocarbons that weigh 0.5 gm at 400 K and 0.821 atmospheric pressure is (3) C2F4.

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