Question

The weight of 350 mL of a diatomic gas at
0°C and 2 atm pressure is 1 gr. The weight in gr
of one atom at NTP is
(A) 2.64 x 10^-23gr
(B) 2.64 x 10^-2-gr
(C) 5.28 x 10^-23gr
(D) 0.82 x 10^-22 gr​

Answer 1

Given Information,

• Volume = 350 ml
• Pressure = 2atm
• Weight of the diatomic gas = 1 gram
• Temperature = 0°C = 273 K

Formulas Used :

Consider the Ideal Gas Equation,

$\sf \: PV = nRT$

According to Mole Concept,

$\sf Moles (n) \leftrightarrow \ No.of \ Molecules \ \leftrightarrow Mass \leftrightarrow Volume \: occupied$

Mathematically,

$\star \: \boxed { \boxed{\sf n = \dfrac{Given \ Mass(M )}{Molecular \ Mass ( W)} = \dfrac{N_o}{N_A} = \dfrac{Volume}{22.4 (STP) }}}$

• $\sf N_o$ is no.of chemical entities and $\sf N_A$ is Avogadro's number.

Gas Constant (R) = 0.08205 L atm/mol K

$\rule{300}{2}$

Now,

$\sf n = \dfrac{RT}{PV} = \dfrac{M}{W}$

Substituting the values,

$\longrightarrow \sf \dfrac{M}{1} = \dfrac{0.08205 \times 273}{2 \times 350 \times 10 {}^{ - 3} } \\ \\ \longrightarrow \boxed{ \boxed{ \sf \: M = 34.36 \: g/mol}}$

Weight of one atom of the gas would be (Number of atoms is 1)

$\because \sf \: \dfrac{Given \ Mass(M )}{Molecular \ Mass ( W)} = \dfrac{N_o}{N_A} \\ \\ \implies \sf \: \dfrac{34.36}{W} = \dfrac{1}{6.02 \times {10}^{23} } \\ \\ \implies \boxed{ \boxed{ \sf \: W_{atom} = 5.7 \times {10}^{ - 23} \: grams}}$

Weight of one atom of the gas at NTP is 5.7 × 10⁻²³ grams.

$\rule{300}{2}$

$\rule{300}{2}$

Answer 2

Given :-

The weight of 350 mL of a diatomic gas at 0°C and 2 atm pressure is 1 gr

To Find :-

The weight in gr of one atom at NTP is

Solution :-

We know that

M = WRT/PV

W - Pressure

R - Universal gas constant

P - Atom given

T - Temperature

V - Volume

1 l = 1000 ml

350 ml = 0.350 ml

M = 1 × 0.0821 × 293/2 × 0.350

M = 293 × 0.0821/0.700

M = 24.0553/0.700

M = 34.36 g/mol

Now

Weight of one atom = 34.36/6.023 × 10²³

Weight = 5.7×10⁻²³ g.

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