Question

calculate the weight of the substance if it's molecular weight is 70 and in the gaseous form occupies 10 lits. at 27℃ and 700mm pressure.

Answer 1

First we should find the volume of gas at S.T.P.

Here P1 = 700 mm, V1 = 10 litre, T1 = 27C = 273 + 27 = 300 K

Also for S.T.P. P2 = 760 mm, V2 =?, T2 = 273 K

Now using ideal gas equation (P1V1)/T1 = (P2V2)/T2

= V2 = ( P1 V1 T2)/( P2 T1)

Putting values we get

V2 = (700 x 10 x 273)/(760 x 300)

= V2 = 8.38 litre (approx.)

Now, as 22.4 litre of any gas at S.T.P. has 1mole (= molecular weight)

= 22.4 litre of gas = 70 g

= 1 litre of gas = 70/22.4

= 8.38 litre of gas = (70/22.4) x 8.38 = 26.187 g (approx.)

= 26.19 g (approx.)

Thus, weight of substance = 26.19 g

Answer 2

Given :-

Initially :

P1 = 700mm

T1=27°C + 273 = 300K

V1 =10 litres

Finally(stp) :

P2 = 760mm

T2 = 273k

To Find :-

V2 = ????

Using the gas equation that is,

=> P1V1 = P2V2

=> 700×10/300 = 760×V2/273

=> 273×70=760×V2×3

=> 19110 = 2280×V2

=> 19110/2280= V2

=> 8.38 litres = V2

Therefore:-

60g of gas X occupies 22.4litres at STP

? g of gas occupies 8.38litres

So:-

=> 22.4 litres =70g

=> 1 litre = 70/22.4g

=> 8.38litres =8.38×70/22.4g =22.45g

Hence:-

V2 = 8.38 litres and weight of substance X is 26.18g.