Question

8gm Methane and 2gm of hydrogen and mixed and kept at 760 mm pressure at 273 K then total volume occupied will be ?
Pls help ​

Answer 1

Given :-

8gm Methane and 2gm of hydrogen and mixed and kept at 760 mm pressure at 273 K

To Find :-

Total volume

Solution :-

Finding moles first

For methane

$\bf\pink{No. \; of \; mole = \dfrac{Total \; mass}{Molar \; mass}}$

$\sf No. \; of \; moles = \dfrac{8}{16}$

$\sf No. \; of \;moles=\dfrac{1}{2}$

$\sf No. \; of \;moles = 0.5$

For hydrogen

$\sf No.\;of\;moles=\dfrac{2}1$

$\sf No. \; of \;moles = 2$

Now

$\bf PV=nRT$

$\sf 1 \times V = 1.5 \times 0.0821 \times 273$

$\sf V = 409.5 \times 0.0821$

$\sf V =33.60\;L$

Answer 2

CONCEPT: Volume is the space occupied by any matter.

• Ideal gas law i.e. PV=nRT

GIVEN: Given mass of methane= 8gm

Given mass of hydrogen= 2gm

Pressure= 760 mm

Temperature= 273 K

FIND: We have to find total volume occupied by the moles of given mixture.

SOLUTION:

no. of moles of methane= 8/16

= 0.5

no. of moles of hydrogen= 2/1

= 2

Total no. of moles in the mixture= 2.5

From ideal gas law: PV = nRT

V = nRT/P

V =2.5×0.0821×273/ 1

V = 56.03 Litres

Hence total volume occupied by the moles of given mixture is 56.03 Litres.

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