A mixture of gas contains 8 g of oxygen and 7 g of nitrogen. To use the equation PV = nRT, the value of n of such mixture will be:

Question

A mixture of gas contains 8 g of oxygen and 7 g of nitrogen. To use the equation PV = nRT, the value of n of such mixture will be: (A)  (1/2)  (B)  (1/4)  (C) 2 (D) 4

Tardigrade Admin · Accepted Answer

Number of moles of oxygen =(8/32)=(1/4) Number of moles of nitrogen =(7/28)=(1/4) Therefore, total number of moles in mixture of two gases, n=(1/4)+(1/4)=(1/2)

Q. A mixture of gas contains $8 g$ of oxygen and $7 g$ of nitrogen. To use the equation $P V = n RT$ , the value of n of such mixture will be:

$\frac{1}{2}$

$\frac{1}{4}$

2

4

Number of moles of oxygen $= \frac{8}{32} = \frac{1}{4}$
Number of moles of nitrogen $= \frac{7}{28} = \frac{1}{4}$
Therefore, total number of moles in mixture of two gases,
$n = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

Q. A mixture of gas contains 8g of oxygen and 7g of nitrogen. To use the equation PV=nRT, the value of n of such mixture will be:

21​

41​

2

4

Number of moles of oxygen =328​=41​ Number of moles of nitrogen =287​=41​ Therefore, total number of moles in mixture of two gases, n=41​+41​=21​

Q. A mixture of gas contains $8 g$ of oxygen and $7 g$ of nitrogen. To use the equation $P V = n RT$ , the value of n of such mixture will be:

$\frac{1}{2}$

$\frac{1}{4}$

Number of moles of oxygen $= \frac{8}{32} = \frac{1}{4}$
Number of moles of nitrogen $= \frac{7}{28} = \frac{1}{4}$
Therefore, total number of moles in mixture of two gases,
$n = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$