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1

### JEE Main 2021 (Online) 27th August Evening Shift

Numerical
100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is x $$\times$$ 10$$-$$2. The value of x is ____________. (Nearest integer)

[Atomic weight : H = 1.008; C = 12.00; O = 16.00]

## Explanation

C3H8(g) + 5O2(g) $$\to$$ 3CO2(g) + 4H2O(l)

t = 0 2.27 mole 31.25 mol
t = $$\infty$$ 0 19.9 mol 6.81 mol 9.08 mol

mole fraction of CO2 in the final reaction mixture (heterogenous)

$${X_{C{O_2}}} = {{6.81} \over {19.9 + 6.81 + 9.08}}$$

= 0.1902 = 19.02 $$\times$$ 10$$-$$2

$$\Rightarrow$$ 19
2

### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
When 10 mL of an aqueous solution of KMnO4 was titrated in acidic medium, equal volume of 0.1 M of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The strength of KMnO4 in grams per litre is __________ $$\times$$ 10$$-$$2. (Nearest integer)

[Atomic mass of K = 39, Mn = 55, O = 16]

## Explanation

Let molarity of KMnO4 = x (Equivalents of KMnO4 reacted) = (Equivalents of FeSO4 reacted)

$$\Rightarrow$$ (5 $$\times$$ x $$\times$$ 10 ml) = 1 $$\times$$ 0.1 $$\times$$ 10 ml

$$\Rightarrow$$ x = 0.02 M

Molar mass of KMnO4 = 158 gm/mol

$$\Rightarrow$$ Strength = (x $$\times$$ 158) = 3.16 g/l
3

### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is ______________. (Nearest integer)

[Atomic mass : Ag = 108, Br = 80]

## Explanation

nAgBr = $${{0.188g} \over {188g/mol}}$$ = 10$$-$$3 mol

$$\Rightarrow$$ nBr = nAgBr = 0.001 mol

$$\Rightarrow$$ massBr = (0.001 $$\times$$ 80) gm = 0.08 gm

$$\Rightarrow$$ mass% = $${{0.08 \times 100} \over {0.2}} = 40\%$$
4

### JEE Main 2021 (Online) 26th August Evening Shift

Numerical
A chloro compound "A".

(i) forms aldehydes on ozonolysis followed by the hydrolysis.

(ii) when vaporized completely 1.53 g of A, gives 448 mL of vapour at STP.

The number of carbon atoms in a molecule of compound A is ___________.

## Explanation

448 ml of A $$\Rightarrow$$ 1.53 gm A

22400 ml of A $$\Rightarrow$$ $${{1.53} \over {448}}$$ $$\times$$ 22400 gm A = 76.5 Agm

$$\mathop {{H_3}CHC - CH - Cl}\limits_{It\,has\,3\,carbon\,atoms} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Zn/{H_2}O}^{{O_3}}} \mathop {C{H_3} - CH = O}\limits_{Aldehyde}$$

& mm is 36 + 5 + 35.5 = 76.5

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