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Test: Mole Concept & Molar Mass (April 7) - JEE MCQ


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10 Questions MCQ Test Daily Test for JEE Preparation - Test: Mole Concept & Molar Mass (April 7)

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Test: Mole Concept & Molar Mass (April 7) - Question 1

For the reaction 2x + 3y + 4z → 5w

Initially if 1 mole of x, 3 mole of y and 4 mole of z is taken. If 1.25 mole of w is obtained then % yield of this reaction is

Detailed Solution for Test: Mole Concept & Molar Mass (April 7) - Question 1

1 moles of x will give = 5/2 = 2.5 mol

Test: Mole Concept & Molar Mass (April 7) - Question 2

Molarity of NaOH in a solution prepared by dissolving 4 g of NaOH in enough water to form 250 ml of solution is:​

Detailed Solution for Test: Mole Concept & Molar Mass (April 7) - Question 2

Weight of NaOH = 4g

Molecular weight of NaOH = 40G

Volume of solution is 250ml

Molarity = ?
M = WMW × 1000 V in ml

M = 440 × 1000250 = 0.4M

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Test: Mole Concept & Molar Mass (April 7) - Question 3

125 ml of 8% w/w NaOH solution (sp. gravity 1) is added to 125 ml of 10% w/v HCl solution. The nature of resultant solution would be ____

Detailed Solution for Test: Mole Concept & Molar Mass (April 7) - Question 3

= 0.25 mole

Test: Mole Concept & Molar Mass (April 7) - Question 4

Ratio  of masses of H2SO4  and Al2 (SO4)3 is grams each containing 32 grams of S is _____

Detailed Solution for Test: Mole Concept & Molar Mass (April 7) - Question 4

Test: Mole Concept & Molar Mass (April 7) - Question 5

18 g of glucose (C6H12O6) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is:​

Detailed Solution for Test: Mole Concept & Molar Mass (April 7) - Question 5

18 g of glucose (C6H12O6) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is 0.1 M.

Test: Mole Concept & Molar Mass (April 7) - Question 6

For the reaction 2A + 3B + 5C → 3D

Initially if 2 mole of A, 4 mole of B and 6 mole of C is taken, With 25% yield, moles of D which can be produced are _____________.

Detailed Solution for Test: Mole Concept & Molar Mass (April 7) - Question 6

Limiting reactant is A
Ideally with 2 moles of A, D formed = 3 moles But yield = 25%
So, moles of D formed = 3 × 0.25 = 0.75 mol

Test: Mole Concept & Molar Mass (April 7) - Question 7

Two elements X (atomic mass = 75) and Y (atomic mass = 16) combine to give a compound having 75.8% of X. The formula of the compound is:

Detailed Solution for Test: Mole Concept & Molar Mass (April 7) - Question 7

So empirical formula of compound = X₂Y₃

For a molecular formula you have to provide molecular mass of the compound.

Test: Mole Concept & Molar Mass (April 7) - Question 8

Equal volumes of 10% (v/v) of HCl is mixed with 10% (v/v) NaOH solution. If density of pure NaOH is 1.5 times that of pure HCl then the resultant solution be :

Detailed Solution for Test: Mole Concept & Molar Mass (April 7) - Question 8

Both have equal volume = V

NaOH mole > HCl mole
Basic Solution

Test: Mole Concept & Molar Mass (April 7) - Question 9

Similar to the % labelling of oleum, a mixture of H3PO4 and P4O10 is labelled as (100 + x) % where x is the maximum mass of water which can react with P4O10 present in 100 gm mixture of H3PO4 and P4O10. If such a mixture is labelled as 127% Mass of P4O10 is 100 gm of mixture, is

Detailed Solution for Test: Mole Concept & Molar Mass (April 7) - Question 9

P4O10 + 6 H2O → 4H3 PO4
284 gm 108gm 392 gm 108 gm water reacts with P4O10 = 284 gm
27 gm water will react with P4O10 = 71 gm

Test: Mole Concept & Molar Mass (April 7) - Question 10

C6H5OH(g) + O2(g) → CO2(g) + H2O(l)

The magnitude of volume change if 30 ml of C6H5OH (g) is burnt with an excess amount of oxygen, is

Detailed Solution for Test: Mole Concept & Molar Mass (April 7) - Question 10

C6H5OH(g)+7O2(g) → 6CO2(g)+3H2O(l)
30 ml
6 x 30= 180 ml of CO2 is produced
Volume used initially  = 30 + 210 = 240
(for C6H5OH)(For O2)
change in volume = 240 - 180 = 60 ml

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