$1 L$ of $N_{2}$ and 7/8 L of $O_{2}$ at the same temperature and pressure were mixed together. what is the relation between the masses of the two gases in the mixture?

M_{N_{2}} = 3 M_{O_{2}}

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M_{N_{2}} = 8 M_{O_{2}}

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M_{N_{2}} = M_{O_{2}}

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M_{N_{2}} = 16 M_{O_{2}}

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Solution

The correct option is D $M_{N_{2}} = M_{O_{2}}$
$P V = \frac{M}{m} R T$

$P \times 1 = \frac{M_{N_{2}}}{28} R T$ ......(1)

$P \times \frac{7}{8} = \frac{M_{O_{2}}}{32} R T$ .......(2)

Dividing equation (i) by (ii), we get

$M_{N_{2}} = M_{O_{2}}$

Hence, the correct option is $C$

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Similar questions

Q. 1.0 litre of

N_{2}

and

\frac{7}{8}

litre of

O_{2}

at the same temperature and pressure were mixed together. What is the relation between the masses of the two gases in the mixture?

Q. Subtract the second expression from the first :

m^{2} n - 8 + m n^{2}

and

7 - m^{2} n - m n^{2}

Q. What is the standard electrode potential for the electrode

M n O {_{4}}^{-} / M n O_{2}

in the solution?
Given:

E {^{o}}_{M n O {_{4}}^{-} / M n^{2 +}} = 1.51 V

and

E {^{o}}_{M n O_{2} / M n^{2 +}} = 1.23 V

Q. Factorize each of the following expressions:
lm² − mn² − lm + n²

While $F e^{3 +}$ is stable, $M n^{3 +}$ is not stabe in acid solution because:
$(M n^{3 +} + e^{-} \to M n^{2 +}, E^{\circ} = 1.5 V; O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V; F e^{3 +} + e^{-} \to F e^{2 +}, E^{\circ} = 0.77 V)$

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1L of N2 and 7/8 L of O2 at the same temperature and pressure were mixed together. what is the relation between the masses of the two gases in the mixture?

The correct option is D MN2=MO2PV=MmRTP×1=MN228RT ......(1)P×78=MO232RT .......(2)Dividing equation (i) by (ii), we getMN2=MO2Hence, the correct option is C

$1 L$ of $N_{2}$ and 7/8 L of $O_{2}$ at the same temperature and pressure were mixed together. what is the relation between the masses of the two gases in the mixture?

The correct option is D $M_{N_{2}} = M_{O_{2}}$
$P V = \frac{M}{m} R T$

$P \times 1 = \frac{M_{N_{2}}}{28} R T$ ......(1)

$P \times \frac{7}{8} = \frac{M_{O_{2}}}{32} R T$ .......(2)

Dividing equation (i) by (ii), we get

$M_{N_{2}} = M_{O_{2}}$

Hence, the correct option is $C$