The weight of $350 m L$ of a diatomic gas at $0^{o} C$ and $2$ atm pressure is $1 g$ . The weight of one atom is:

Open in App

Solution

From Ideal gas equation- $P V = n R T$

Here,
$P = 2 a t m$ , $V = 350 m l = 0.35 l$
$R = 0.0821 l a t m / k m o l$ and $T = 0^{o} C = 273 K$

Now, $n = \frac{P V}{R T}$

$n = \frac{2 \times 0.35}{0.0821 \times 273}$

$n = 0.0312 m o l$

Now, the weight of 0.0312 mol of gas is 1 gram

Therefore, the weight of 1mol of gas would be $= \frac{1}{0.0312} = 32 g / m o l$

Since the gas is diatomic therefore weight of 1 mole atom $= 16 g / m o l$

Suggest Corrections

Similar questions

Q. The weight of

350

mL of a diatomic gas at

0^{o} C

and

2

atm pressure is 1 g. The weight in g of one atom at

N T P

is:

Q. At

0^{o} C

and 2 atm pressure, the volume of 1 gram of a diatomic gaseous element is 350

m l

. Weight of 1 atom of the element in grams is:

One litre of a gas weights 2g at 300 K and 1 atm. pressure. If the pressure is made 0.75 atm, at which temperature will one litre of the same gas weigh 1g?

Q. The weight of

181.60 mL

of a diatomic gas at

0^{\circ} C

and

1 bar

pressure is

1 g

. The mass of one molecule (in g) of the gas is:

(N_{A}

is the Avogadro's constant)

Q. One liter of a gas weight

4 g

300 K

and

1

atm. If the pressure is reduced to

0.75

atm,the temperature at which one litre of the same gas weight

2 g

is:

Join BYJU'S Learning Program

The weight of 350mL of a diatomic gas at 0oC and 2 atm pressure is 1g. The weight of one atom is:

The weight of $350 m L$ of a diatomic gas at $0^{o} C$ and $2$ atm pressure is $1 g$ . The weight of one atom is: