Question

A hydrocarbon CxHy has mass ratio between hydrogen and carbon 1:10.5. One litre of the
hydrocarbon at 127°C and 1 atm measure weights 2.89. Find the molecular formula of the
hydrocabon.
please solve it in detail​

Answer 1

Question

A hydrocarbon CxHy has mass ratio between hydrogen and carbon 1:10.5. One litre of the

A hydrocarbon CxHy has mass ratio between hydrogen and carbon 1:10.5. One litre of thehydrocarbon at 127°C and 1 atm measure weights 2.89. Find the molecular formula of the

A hydrocarbon CxHy has mass ratio between hydrogen and carbon 1:10.5. One litre of thehydrocarbon at 127°C and 1 atm measure weights 2.89. Find the molecular formula of thehydrocabon.

Answer⬇️⬇️

As we know,PV=nRT= Mw RTor 1= M2.8 ×0.0821×400or M=91.95≈92Let the formula be C a H b . So 12a+b=92b12a =10.5(given)11.5b=92or b=8∴ a=7Therefore, the formula is C

As we know,PV=nRT= Mw RTor 1= M2.8 ×0.0821×400or M=91.95≈92Let the formula be C a H b . So 12a+b=92b12a =10.5(given)11.5b=92or b=8∴ a=7Therefore, the formula is C 7

As we know,PV=nRT= Mw RTor 1= M2.8 ×0.0821×400or M=91.95≈92Let the formula be C a H b . So 12a+b=92b12a =10.5(given)11.5b=92or b=8∴ a=7Therefore, the formula is C 7

As we know,PV=nRT= Mw RTor 1= M2.8 ×0.0821×400or M=91.95≈92Let the formula be C a H b . So 12a+b=92b12a =10.5(given)11.5b=92or b=8∴ a=7Therefore, the formula is C 7 H

Answer 2

Explanation:

As we know,

$PV = nRT = \frac{w}{M}RT$

$or 1 = \frac{2.8}{M} \times 0.0821 \times 400$

or M= 91.95~92

Let the formula be CaHb.

So 12a + b= 92

$\frac{12a}{b} = 10.5 (given)$

11.5b=92

or b= 8

Therefore, a= 7

Therefore, the formula is C7H8

molar mass= 12*7+1*8= 92gm/mole

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