NEET

The number of molecules in 16 g of methane is

3.0 × 10₂₃
\(\frac{16}{6.02}\) × 10²³
6.023 × 10₂₃
\(\frac{16}{3.0}\) × 10²³

Solution

16 g CH₄ is 1 mol. Hence number of molecules= Avogadro number = 6.023 × 10²³

3 g of an oxide of a metal is converted to chloride completely and it yielded 5 g of chloride. The equivalent weight of the metal is

3.325
33.25
12
20

Solution

7.5 grams of a gas occupy 5.6 litres of volume at STP. The gas is

N₂O
NO
CO
CO₂

Solution

PV = nRT ∴ 5.6 × 1 = \(\frac{7.5}{M.Wt}\) × 0.0821 × 273

M. Wt = 30.12 Hence gas NO

How many moles of magnesium phosphate, Mg₃(PO₄)₂ will contain 0.25 mole of oxygen atoms?

1.25 × 10^–2
2.5 × 10^–2
0.02
3.125 × 10^–2

Solution

1 Mole of Mg₃(PO₄)₂ contains 8 mole of oxygen atoms
∴8 mole of oxygen atoms = 1 mole of Mg₃(PO₄)₂
0.25 mole of oxygen atom = \(\frac{1}{8}\) × 0.25 mole of Mg₃(PO₄)
=3.125 × 10^-2mole of Mg₃(PO₄)₂

Number of atoms in 558.5 gram Fe (at. wt. of Fe = 55.85g mol^–1) is

twice that in 60 g carbon
6.023 × 10²²
half that in 8 g He
558.5 × 6.023 × 10²³

Solution

Fe (no. of moles)= \(\frac{558.8}{55.85}\)= 10 moles = 10 N_A atoms.
No. of moles in 60 g of C = 60/12 = 5 moles =5 N_A atoms.

Which has maximum number of molecules?

7 gm N₂
2 gm H₂
16gm NO₂
16gm O₂

Solution

2g of H₂ means one mole of H₂, hence contains 6.023 × 10²³ molecules. Others have less than one mole,so have less no.of molecules.

The number of gram molecules of oxygen in 6.02 × 10²⁴ CO molecules is

10 gm molecules
5 gm molecules
1 gm molecules
0.5 gm molelcules

Solution

6.02 × 10²³ molecules of CO=1 mole of CO 6.02 × 10²⁴CO molecules = 10 moles CO
=10g atoms of O= 5 g molecules of O₂

Calculate the volume at STP occupied by 240 gm of SO₂.

64
84
59
73

Solution

Molecular weight of SO₂= 32 + 2 ×16 = 64
64 g of SO₂ occupies 22.4 litre at STP
240 g of SO₂ occupies = \(\frac{22.4}{64}\) × 240 = 84 litre at STP

What is the mass of 1 molecule of CO.

2.325 × 10^–23
4.65 × 10^–23
3.732 × 10^–23
2.895 × 10^–23

Solution

Gram molecular weight of CO = 12 + 16 = 28 g
6.023× 10²³ molecules of CO weight 28g
1 molecule of CO weighs= \(\frac{28}{6.02 \times 10^{23}}\) = 4.65 × 10^-23

The vapour density of a gas is 11.2, then 11.2 g of this gas at N.T.P. will occupy a volume-

11.2 L
22.4 L
11.2 mL
22.4 mL

Solution

Molecular mass of any gas occupies 22.4 L at N.T.P.
Vapour density = \(\frac{Molecular Mass}{2}\)
Vapour density of any gas occupies a volume of 11.2 litres at N.T.P

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