The mass of 350- cm - 3 - of a diatomic gas 273 K 2 atmospheres pressure is one gram- Calculate the mass of one atom of the gas-5-3times 10-23-g2-65times 10-23- g12-02times 10-23-gNone of these

Question

Toppr Admin · Accepted Answer

N-PVRT-2-xD7-0-350-082-xD7-273-x2248-0-031mol0-031 mol weight-xA0- -xA0- -xA0- -xA0- -xA0- -xA0-1g0-031-xD7-6022-xD7-1023atoms-xA0- -xA0- -xA0- -xA0- -xA0- 1g-xA0- -xA0- -xA0- -xA0- -xA0- -xA0-10-031-xD7-6-022-xD7-10-x2212-23-xA0- -xA0- -xA0- -xA0- -xA0- -xA0-5-3-xD7-10-x2212-23g

The mass of $350 {c m}^{3}$ of a diatomic gas at 273 K at 2 atmospheres pressure is one gram. Calculate the mass of one atom of the gas.
$5.3 \times 10^{- 23} g$
$2.65 \times 10^{- 23} g$
$12.02 \times 10^{23} g$
None of these

$n = \frac{P V}{R T}$
$= \frac{2 \times 0.35}{0.082 \times 273}$
$\approx= 0.031 m o l$
$0.031$ mol weight $1 g$
$0.031 \times 6022 \times 10^{23} a t o m s$ = $1 g$
$= \frac{1}{0.031 \times 6.022} \times 10^{- 23}$
$= 5.3 \times 10^{- 23} g$

The mass of 350cm3 of a diatomic gas at 273 K at 2 atmospheres pressure is one gram. Calculate the mass of one atom of the gas.5.3×10−23g2.65×10−23g12.02×1023gNone of these

n=PVRT=2×0.350.082×273≈=0.031mol0.031 mol weight 1g0.031×6022×1023atoms= 1g =10.031×6.022×10−23 =5.3×10−23g

The mass of $350 {c m}^{3}$ of a diatomic gas at 273 K at 2 atmospheres pressure is one gram. Calculate the mass of one atom of the gas.
$5.3 \times 10^{- 23} g$
$2.65 \times 10^{- 23} g$
$12.02 \times 10^{23} g$
None of these

$n = \frac{P V}{R T}$
$= \frac{2 \times 0.35}{0.082 \times 273}$
$\approx= 0.031 m o l$
$0.031$ mol weight $1 g$
$0.031 \times 6022 \times 10^{23} a t o m s$ = $1 g$
$= \frac{1}{0.031 \times 6.022} \times 10^{- 23}$
$= 5.3 \times 10^{- 23} g$