How many of gms nitrogen are in $46 g$ of dinitrogen tetroxide?

92 g

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15 g

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7.0 g

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14 g

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Solution

The correct option is D $14 g$
Percentage of Nitrogen by mass in $N_{2} O_{4} = \frac{2 \times 14}{2 \times 14 + 4 \times 16} \times 100 = 30.43 %$
So, mass of Nitrogen in $46 g$ of $N_{2} O_{4} = (% of Nitrogen) \times (mass of sample) = (\frac{30.43}{100}) \times (46) = 14 g$

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Similar questions

Which of the following expressions is the correct equilibrium-constant expression for the equilibrium between dinitrogen tetroxide and nitrogen dioxide?

N₂O₄ (g) ⇌ 2NO₂ (g)

Suppose the reaction is carried out starting with

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Q. At

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, dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one temperature.

Q. At

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, dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

Q. When a vapour, at atmospheric pressure was gradually heated from

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it's colour was found to deepen at first and then fade as the temperature was raised above

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. At

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, the vapour was almost colourless, but its colour deepened when the pressure was raised at this temperature. The vapour was:

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How many of gms nitrogen are in 46g of dinitrogen tetroxide?

The correct option is D 14 gPercentage of Nitrogen by mass in N2O4=2×142×14+4×16×100=30.43%So, mass of Nitrogen in 46 g of N2O4=(% of Nitrogen)×(mass of sample)=(30.43100)×(46)=14 g

How many of gms nitrogen are in $46 g$ of dinitrogen tetroxide?

The correct option is D $14 g$
Percentage of Nitrogen by mass in $N_{2} O_{4} = \frac{2 \times 14}{2 \times 14 + 4 \times 16} \times 100 = 30.43 %$
So, mass of Nitrogen in $46 g$ of $N_{2} O_{4} = (% of Nitrogen) \times (mass of sample) = (\frac{30.43}{100}) \times (46) = 14 g$