The average molecular mass of a mixture of gas containing nitrogen and carbon dioxide is 36 . The mixture contain 280 g of nitrogen, therefore, the amount of CO 2 present in the mixture is

Question

The average molecular mass of a mixture of gas containing nitrogen and carbon dioxide is 36 . The mixture contain 280 g of nitrogen, therefore, the amount of CO 2 present in the mixture is (A) 440 g (B) 44 g (C) 0.1 mol (D) 880 g

Tardigrade Admin · Accepted Answer

Let amount of CO 2 persent = x g 36=((280/28) × 28+(x/44) × 44/(280)28+(x)44 36=(280+x/10+x / 44)=((280+x) × 44/440+x) or 15840+36 x=12320+44 x or 8 x=3520 x=(3520/8)=440 g

Q. The average molecular mass of a mixture of gas containing nitrogen and carbon dioxide is $36.$ The mixture contain $280 g$ of nitrogen, therefore, the amount of $C O_{2}$ present in the mixture is

440 g

44 g

0.1 mol

880 g

Let amount of $C O_{2}$ persent $= x g$

$36 = \frac{\frac{280}{28} \times 28 + \frac{x}{44} \times 44}{\frac{280}{28} + \frac{x}{44}}$

$36 = \frac{280 + x}{10 + x /44} = \frac{( 280 + x ) \times 44}{440 + x}$

or $15840 + 36 x = 12320 + 44 x$

or $8 x = 3520$

$x = \frac{3520}{8} = 440 g$

Q. The average molecular mass of a mixture of gas containing nitrogen and carbon dioxide is 36. The mixture contain 280g of nitrogen, therefore, the amount of CO2​ present in the mixture is