Molecular density of N2 at 1.00 atm and 25°C is (assume ideal behaviour)

Question

Molecular density of N2 at 1.00 atm and 25°C is (assume ideal behaviour) (A) 2.46 × 1022 dm -3 (B) 2.46 × 1025 m -3 (C) 2.40 × 1019 cm -3 (D) All of these

Tardigrade Admin · Accepted Answer

Molecular density is the number of molecular per unit volume.

Let, moles

= n

, Volume

= V L

p V = n RT

\frac{n}{V} = \frac{p}{RT}

= \frac{1 a t m}{0.0821 a t m m o l ^{- 1} K ^{- 1} \times 298 K}

= \frac{1}{0.0821 \times 298} m o l L^{- 1}

\frac{n N _{0}}{V}

(molecular density)

= \frac{6.02 \times 1 0 ^{23}}{0.0281 \times 298}

= 2.46 \times 1 0^{22} d m^{- 3}

= 2.46 \times 1 0^{25} m^{- 3}

or

= 2.46 \times 1 0^{19} c m^{- 3}

Q. Molecular density of $N_{2}$ at $1.00 a t m$ and 25°C is (assume ideal behaviour)

$2.46 \times 1 0^{22} d m^{- 3}$

$2.46 \times 1 0^{25} m^{- 3}$

$2.40 \times 1 0^{19} c m^{- 3}$

All of these

Q. Molecular density of N2​ at 1.00atm and 25°C is (assume ideal behaviour)

2.46×1022dm−3

2.46×1025m−3

2.40×1019cm−3

All of these

Molecular density is the number of molecular per unit volume. Let, moles=n, Volume =VL pV=nRT Vn​=RTp​ =0.0821atmmol−1K−1×298K1atm​ =0.0821×2981​molL−1 VnN0​​ (molecular density) =0.0281×2986.02×1023​ =2.46×1022dm−3 =2.46×1025m−3 or =2.46×1019cm−3

Q. Molecular density of $N_{2}$ at $1.00 a t m$ and 25°C is (assume ideal behaviour)

$2.46 \times 1 0^{22} d m^{- 3}$

$2.46 \times 1 0^{25} m^{- 3}$

$2.40 \times 1 0^{19} c m^{- 3}$