1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Morning Slot

NaH is an example of :
A
metallic hydride
B
saline hydride
C
electron-rich hydrid
D
molecular hydride

Explanation

NaH is an example of ionic hydride which is also known as saline hydride.
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 12th January Morning Slot

The hardness of a water sample (in terms of equivalents of CaCO3) containing 10–3 M CaSO4 is (Molar mass of CaSO4 = 136 g mol–1)
A
90 ppm
B
100 ppm
C
50 ppm
D
10 ppm

Explanation

nCaSO4 $$ \times $$ Van't hoff factor = nCaCO3 $$ \times $$ Van't hoff factor

$$ \Rightarrow $$ 10-3 $$ \times $$ 2 = nCaCO3 $$ \times $$ 2

$$ \Rightarrow $$ nCaCO3 = 10-3 mol in 1 L

$$ \therefore $$ wCaCO3 = 100 $$ \times $$ 10-3 g CaCO3 in 1 L solution

$$ \therefore $$ hardness in terms of CaCO3

= $${{{w_{CaC{O_3}}}} \over {{w_{Total}}}} \times {10^6}$$

= $${{100 \times {{10}^{ - 3}}} \over {1000}} \times {10^6}$$

= 100 ppm
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 12th January Evening Slot

The volume strength of 1M H2O2 is (Molar mass of H2O2 = 34 g mol–1)
A
5.6
B
22.4
C
11.35
D
16.8

Explanation

At   STP,

Volume strength = 11.35 $$ \times $$ Molarity

$$ \therefore $$  Volume strength of H2O2 = 11.35 $$ \times $$ 1 = 11.35
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 8th April Morning Slot

100 mL of a water sample contains 0.81 g of calcium bicrabonate and 0.73 g of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of CaCO3 is : (Molar mass of calcium bicarbonate is 162 g mol–1 and magnesium bicarbonate is 146 g mol–1)
A
1,000 ppm
B
100 ppm
C
10,000 ppm
D
5,000 ppm

Explanation

Here hardness of water is expressed in terms of CaCO3.

$$ \therefore $$ Equivalent of CaCO3

= Equivalent of Ca(HCO3)2 + Equivalent of Mg(HCO3)2

$$ \Rightarrow $$ 2 $$ \times $$ $${W \over {100}}$$ = $${{0.81} \over {162}} \times 2 + {{0.73} \over {146}} \times 2$$

$$ \Rightarrow $$ W = 1 gm

Volume of water = 100 mL

$$ \therefore $$ Mass of water = 100 g

$$ \therefore $$ Hardness = $${{1.0} \over {100}} \times {10^6}$$ = 10000 ppm

Questions Asked from Hydrogen

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