JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Surface Chemistry

3.12 g of oxygen is adsorbed on 1.2 g of platinum metal. The volume oxygen adsorbed per gram of the adsorbent at 1 atm and 300 K in L is ______

[R = 0.0821 L atm K^{- 1} mole^{- 1}]

Answer: 2

Solution

The amount of oxygen adsorbed is

n = \frac{3.12}{32} = 0.0975 mol

1

atm and

300

K this gas occupies

V = \frac{n R T}{P} = \frac{0.0975 \times 0.0821 \times 300}{1} = 2.4 L

The adsorbent is

1.2

g of platinum, so the volume adsorbed for every gram of adsorbent is

\frac{2.4}{1.2} = 2 L g^{- 1}

The volume of oxygen adsorbed per gram of the adsorbent is

2

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Integer answer