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1

JEE Main 2021 (Online) 25th February Morning Slot

Numerical
In basic medium $$Cr{O_4}^{2 - }$$ oxidises $${S_2}{O_3}^{2 - }$$ to form $$S{O_4}^{2 - }$$ and itself changes into $$Cr{(OH)_4}^ - $$. The volume of 0.154 M $$Cr{O_4}^{2 - }$$ required to react with 40 mL of 0.25 M $${S_2}{O_3}^{2 - }$$ is __________ mL. (Rounded off to the nearest integer)
Your Input ________

Answer

Correct Answer is 173

Explanation

$$17{H_2}O + 8Cr{O_4} + 3{S_2}{O_3}\buildrel {} \over \longrightarrow 6S{O_4} + 8Cr{(OH)_4}^ - + 2O{H^ - }$$

Applying mole-mole analysis

$${{0.154 \times v} \over 8} = {{40 \times 0.25} \over 3}$$

v = 173 mL
2

JEE Main 2021 (Online) 24th February Evening Slot

Numerical
The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O2 for complete oxidation and produces 4 times its own volume of CO2 is CxHy. The value of y is _____________.
Your Input ________

Answer

Correct Answer is 8

Explanation

Combustion reaction :

$${C_x}{H_y}(g) + \left( {x + {y \over 4}} \right){O_2}(g) \to xC{O_2}(g) + {y \over 2}{H_2}O(l)$$

Suppose, volume of CxHy is V and volume of O2 is 6 times greater than CxHy = 6V

then volume of xCO2 $$\Rightarrow$$ Vx = 4 V

x = 4

Since, $${V_{{O_2}}} = 6 \times {V_{{C_x}{H_y}}}$$

$$V\left( {x + {y \over 4}} \right)$$ = 6V

$$\left( {x + {y \over 4}} \right) = 6$$ ..... (i)

Put value of x = 4 in Eq. (i) we get,

$$4 + {y \over 4} = 6 \Rightarrow y = 8$$
3

JEE Main 2021 (Online) 24th February Morning Slot

Numerical
4.5 g of compound A (MW = 90) was used to make 250 mL of its aqueous solution. The molarity of the solution in M is x $$ \times $$ 10-1.
The value of x is _______. (Rounded off to the nearest integer)
Your Input ________

Answer

Correct Answer is 2

Explanation

Given, weight of compound A = 4.5 g

Molecular weight of compound A = 90 g/mol

Volume of solution (in mL) = 250 mL

Now, molarity is defined as number of moles of solute or compound A divided by volume of solution (in L).

$$M = {{Number\,of\,moles\,of\,solute\,(n)} \over {Volume\,of\,solution}}$$

$$ = {{{{4.5} \over {90}}} \over {{{250} \over {1000}}}} = 0.2$$ = 2 $$\times$$ 10$$-$$1 M

$$\therefore$$ $$n = {{Weight\,of\,solute\,(compound\,A)} \over {Molecular\,weight\,of\,solute\,(compound\,A)}}$$

Hence, x $$\times$$ 10$$-$$1 $$\mu$$

x = 2
4

JEE Main 2020 (Online) 6th September Morning Slot

Numerical
In an estimation of bromine by Carius method, 1.6 g of an organic compound gave 1.88 g of AgBr. The mass percentage of bromine in the compound is _______.
(Atomic mass, Ag = 108, Br = 80 g mol–1)
Your Input ________

Answer

Correct Answer is 50

Explanation

Mass of organic compound = 1.6 gm

Mass of AgBr = 1.88 gm

Moles of Br = Moles of AgBr = $${{1.88} \over {188}}$$ = 0.01

Mass of Br = 0.01 × 80 = 0.80 gm

% of Br = $${{0.80 \times 100} \over {1.60}}$$ = 50 %

Questions Asked from Some Basic Concepts of Chemistry

On those following papers in Numerical
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