Question

the mole percent of oxygen present in gaseous mixture containing 14 g of nitrogen and 32 g of o oxygen is?​

Answer 1

Answer:

50%of oxygen

Explanation:

1 mole of nitrogen =14/14=1

1............... oxygen=32/32=1

so mole% of oxygen is 50%

Answer 2

Answer:

The correct answer is $66.66 \%$.

Explanation:

A mole exists as a way of calculating collections of discrete chemical objects such as atoms or molecules.

Given the mass of nitrogen $=14 \mathrm{~g}$

The molar mass of nitrogen gas $=28 \mathrm{~g}$

Given the mass of oxygen $=32 \mathrm{~g}$

The molar mass of oxygen gas $=32 \mathrm{~g}$

Number of moles of oxygen in the mixture,

$X_{1}=\frac{\text { massofoxygen }}{\text { molarmassofoxygen }}$

$=\frac{32}{32}=1$

Number of moles of Nitrogen in the mixture,

$X_{2}=\frac{\text { massofnitrogen }}{\text { molemassofnitrogen }}$

$=\frac{14}{28}=0.5$

The mole percent of oxygen

$=\frac{x_{1}}{x_{1}+x_{2}} \times 100$

$=\frac{1}{1+0.5} \times 100$

$=\frac{100}{1.5}=66.66 \%$

Therefore, the correct answer is $66.66 \%$.

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