JEE Advanced Mix Test 38

Result

JEE Advanced Mix Test 38

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Weekly Quiz Competition

Question 1
4 / -1

The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation

When a certain conductivity cell (C) was filled with 25 x 10^–4 (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl– and SO4^–2 are 80 ohm–1 cm²mole–1 and 160 ohm^–1 cm² mole^–1respectively.

What is the molar conductance of NaCl at infinite dilution ?

A
147 ohm^–1 cm² mole^–1

B
107 ohm^–1 cm² mole^–1

C
127 ohm^–1 cm² mole^–1

D
157 ohm^–1 cm² mole^–1

Solution
Question 2
4 / -1

The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation

When a certain conductivity cell (C) was filled with 25 x 10^–4 (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl^– and SO4^–2 are 80 ohm^–1cm² mole–1 and 160 ohm–1 cm² mole–1 respectively. Q. What is the cell constant of the conductivity cell (C) ?

A
0.385 cm^–1

B
3.85 cm^–1

C
38.5 cm^–1

D
0.1925 cm^–1

Solution
Question 3
4 / -1

The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation

When a certain conductivity cell (C) was filled with 25 x 10^–4 (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl– and SO4^–2 are 80 ohm^–1 cm²mole^–1 and 160 ohm^–1 cm² mole–1 respectively. Q. If the cell (C) is filled with 5 x 10^–3(N) Na₂SO₄the obserbed resistance was 400 ohm. What is the molar conductance of Na₂SO₄ ?

A
19.25 ohm^–1 cm² mole^–1

B
96.25 ohm^–1 cm² mole^–1

C
385 ohm^–1 cm² mole^–1

D
192.5 ohm^–1 cm² mole^–1

Solution
Question 4
4 / -1

Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociation where n is the number of products (ions or molecules) obtained per mole of the reactant & i = for association.

Where n is number of reactant particles associated to give 1 mole product. A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant Kb. The solute dimerises in the solution according to the following equation. The degree of association is α.

The vant Hoff factor will be [ if we start with one mole of X ]

A

B

C

D

Solution
Question 5
4 / -1

A
(A) → q ; (B) → p ; (C) → r, s, t ; (d) → s, r, t

B
(A) → s, r,t ; (B) → r, s,t ; (C) → p ; (d) → q

C
(A) → p ; (B) → q ; (C) → s, q ; (d) → p, q

D
(A) → p,q ; (B) → q, r ; (C) → s, p ; (d) → r, s

Solution
Question 6
4 / -1

A
(A) → p,q,t ; (B) → s,t ; (C) → p, r, s,t ; (D) → p,q,t

B
(A) → s,t ; (B) → p,q ; (C) → p, t ; (D) → r,s,t

C
(A) → p,q,t ; (B) → p,r,s,t ; (C) → s,t ; (D) → p,q,t

D
(A) → q,r ; (B) → t ; (C) → q, s, ; (D) → q

Solution

(A) First order :– t_1/2 is constant. (p)

concentration and rate decrease exponentially.(q)

Arhennius equation is applicable for every order reaction (t)

(B) Zero order :- Rate is constant. (p)

Concentration decreases linearly with t. (r)
Question 7
4 / -1

Two roads OA and OB intersect at an angle of 60º. A car driver approaches O from A where OA = 800m at a uniform speed of 20m/s. Simultaneously another car moves from O towards B at a uniform speed of 25m/s. If t is time when two cars are closest, find t.

A
1

B
2

C
1040/61

D
3/2

Solution

Let the distance between two cars after t seconds be s. Distance covered by first car in t second is 20t meter, so its distance from O will be (800 – 20t) meter and distance covered by second car will be 25t meter.
Question 8
4 / -1

If f(x) satisfies x + |f (x)| = 2f (x) then f^–1 (x) satisfies

A
3x + |f^–1 (x)| = 2 f ^–1 (x)

B
x + |f^–1 (x)| = 2 f ^–1 (x)

C
f ^–1 (x) – |x| = 2x

D
3x – |f ^–1 (x)| = 2 f ^–1 (x) where f(x) > 0

Solution
Question 9
4 / -1

A

B

C

D
None of these

Solution
Question 10
4 / -1

Let f : R → R and g : R → R be two one - one and onto functions such that they are mirror images of each other about the line y = 0, then h (x) = f(x) + g(x) is

A
one - one and onto

B
one - one but not onto

C
not one - one but onto

D
None of these

Solution

f : R → R & g : R →R be two one-one onto functions such that f & g are mirror images of each other about line y = 0. It means one is –ve of the other
i.e. f(x) = – g(x)
⇒ f(x) + g(x) = 0
⇒ h(x) = 0

h(x) is not onto as well as not one-one