This question was given in your selina concise chemistry part-2 for class 10 ch-5 mole concept ex 5B Q 26 but you all are doing it wrong and that's why we have to suffer. Now please for god's sake give the correct answer to this question!!!!!!

Asked by Supratim Sircar | 14th May, 2014, 02:33: PM

Expert Answer:

Given:

P= 1140 mm Hg

Density = D = 2.4 g / L

T = 273 0C = 273+273 = 546 K

M = ?

We know that, at STP, the volume of one mole of any gas is 22.4 L

Hence we have to find out the volume of the unknown gas at STP.

First apply Charle’s law.

We have to find out the volume of one liter of unknown gas at standard temperature 273 K

V1= 1 L T1 = 546 K

V2=? T2 = 273 K

V1/T1 = V2/ T2

V2 = (V1 x T2)/T1

= (1 L x 273 K)/546 K

= 0.5 L

We have found out the volume at standard temperature. Now we have to find out the volume at standard pressure.

Apply Boyle’s law.

P 1 = 1140 mm Hg V1 = 0.5 L

P2 = 760 mm Hg V2 = ?

P1 x V1 = P2 x V2

V2 = (P1 x V1)/P2

= (1140 mm Hg x 0.5 L)/760 mm Hg

= 0.75 L

Now, 22.4 L is the volume of 1 mole of any gas at STP, then 0.75 L is the volume of X moles at STP

X moles = 0.75 L / 22.4 L

= 0.0335 moles

The original mass is 2.4 g

n = m / M

0.0335 moles = 2.4 g / M

M = 2.4 g / 0.0335 moles

M= 71.6 g / mole

Hence, the gram molecular mass of the unknown gas is 71.6 g.

Answered by Prachi Sawant | 15th May, 2014, 11:14: AM

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