RC Mukherjee Test: Mole Concept - JEE MCQ

Molecular mass = 2 x vapour density
= 2 x 22
= 44
Therefore, molecular mass of a gas will be 44.

Molecular Mass of Ethanol
= 2(12) + 5 (1) + 16 + 1 
= 24 + 5 + 17 
= 46
Now,
Moles (n) = Given Mass (m)/Molecular Mass (M)
Thus, to = nM
= (0.2) (46)
= 9.2

In the equation both side of reaction moles of atoms is conserved ie, both reactant and product contain 4H and 2O but there is 3 (2+1) moles of reactant molecules turns into 2 moles of product molecules. 

C = 84/12 = 7 mole

H₂ = 12 g = 6 mole

O₂ = 56/22.4 = 5/2 mole

12C + 11H₂ + 11/2 O₂ → C₁₂H₂₂O₁₁

L.R. = O₂

11/2 mole O2 produce 1 mole sucrose 5/2 mole O2 will for 5/11 mole sucrose mass of sucrose = 5/11 × (mol. mass)

= 5/11 × 342

= 155.45 g

Let atomic weight of x = Mx

atomic weight of y = M_y

we know,

mole = weight /atomic weight

a/c to question,

mole of xy₂ = 0.1

so,

0.1 = 10g/( M_x +2M_y)

M_x + 2M_y = 100g -------(1)

for x₃y₂ ; mole of x₃y₂ = 0.05

0.05 = 9/( 3M_x + 2M_y )

3M_x + 2M_y = 9/0.05 = 9× 20 = 180 g ---(2)

solve eqns (1) and (2)

2M_x = 80

M_x = 40g/mol

and M_y = 30g/mole

Gram atomic mass of helium is 4 g so number of atoms present in 52 g of helium
52/4 × N_A = 13 N_A

As vapour density of mixture is 25,

Molar mass of mixture is 2×25=50.

Now, assume mole % of B in mixture is x.

Mass of 1 mole of Cl₂ is 71 g. Hence mass of 4.50 moles is 71 X 4.50 = 319.5 g

Mass of one mole of Br₂  is 160g. Hence mass of four moles of molecular bromine is 4 X 160 = 640 g.

Given weight of N₂ gas = 1.4 g

Molar mass of N₂ gas = 28 g

So, mole = given mass/ molar mass

⇒ mole = 1.4/28 = 1/20 mole

Now, number of molecules = mole × avogadro number

⇒ number of molecules = 1/20 × 6.022 × 10²³

⇒ number of molecules = 3.011 × 10²²

Now, we are asked for number of atoms. In N₂, there are 2 atoms, so to obtain number of atoms we will multiply with 2 in number of molecules.

⇒ number of atoms = 2 × 3.011 × 10²²

⇒ number of atoms = 6.022 × 10²²

Molar mass of NH₃ is 17g. So the volume of 17g of NH₃ i.e 1 mole NH₃ at N.T.P. will be 22.4L.

1g hydrogen has the maximum number of moles because atomic mass of hydrogen is smallest among all.

One molecule of F₂  contains 2 atoms of fluorine.

1 mol of fluorine contains 2 X 6.023 X 10²³ atoms

So, 0.147 moles will contain 0.147 X 2 X 6.023 X 10²³ = 1.76 X 10²³ atoms of flourine.

1 mole of a substance is equal to its molecular mass expressed in terms of grams.
Gram molecular mass of CO₂ = 12 + (16 Χ 2) = 12 + 32 = 44 g
Mass of 6.023 X10₂₂ molecules of CO₂ is 44g/10 = 4.4g.