Hydrogen Test

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Hydrogen Test - 60

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Weekly Quiz Competition

Question 1
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Which of the following represents the chemical equation involved in the preparation of $$H_2O_2$$ from barium peroxide?

A
$$BaO_2.8H_2O+H_2SO_4\rightarrow BaSO_4+H_2O_2+8H_2O$$

B
$$CH_3CHOHCH_3\rightarrow CH_3COOCH_3+H_2O_2$$

C
$$BaO_2+CO_2+H_2O\rightarrow BaCO_3+H_2O_2$$

D
$$Ba_3(PO_4)_2+3H_2SO_4\rightarrow 3BaSO_4+2H_3PO_4$$

Solution

Barium peroxide react with sulfuric acid to produce Hydrogen peroxide and Barium sulfate.

Isopropyl Alcohol oxidises to Acetone and water, and not hydrogen peroxide.

Barium peroxide react with $$CO_{2}$$ carbonic acid. It then racts with barium to form barium carbonate and hydrogen peroxide. Barium carbonate is insoluble and is easy to remove by filtration. This process is widely used to manufacture $$H_{2}O_{2}$$.

Barium phosphate reacts with Sulphuric acid to form Barium sulfate and Phosphoric acid.

The correct option is C.
Question 2
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Metal hydrides are ionic, covalent or molecular in nature. Among $$LiH,\ NaH,\ KH,\ RbH$$ and $$ CsH$$, the correct order of increasing ionic character is:

A
$$LiH > NaH > CsH > KH > RbH$$

B
$$LiH < NaH < KH < RbH < CsH$$

C
$$RbH > CsH > NaH > KH > LiH$$

D
$$NaH > CsH > RbH > LiH > KH$$

Solution

Ionic character increases with the size of the cation as we move down the group. As the ionisation enthalpy decreases on moving from top to bottom in a group, the tendency to lose electron by metal increases. Hence ionic character increases.
Question 3
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Two structure of $$H_2O_2$$are drawn below. Identify the phase $$(X)$$ and $$(Y)$$ of $$H_2O_2$$.

A
$$(X)$$ is the structure of $$H_2O_2$$ in gas phase and $$(Y)$$ in solid phase

B
$$(X)$$ is structure of $$H_2O_2$$ in solid phase and $$(Y)$$ in gas phase

C
$$(X)$$ and $$(Y)$$ are structures of $$H_2O_2$$ in gas phase

D
$$(X)$$ and $$(Y)$$ are structures of $$H_2O_2$$ in solid phase

Solution

$$(X)$$ is the structure of $$H_2O_2$$ in the gas phase and $$(Y)$$ in the solid phase.

$$O-O$$ single bond length is higher in the gas phase. $$H-O-O$$ bond angle is higher in the solid phase.

Hence, the correct answer is option $$\text{A}$$.
Question 4
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$$ 5.0 cm^3 of H_2O_2 $$ liberates 0.508 g of iodine from an acidified KI solution. The strength of $$H_2O_2$$ solution in terms of volume strength at STP is:

A
6.48 volumes

B
4.48 volumes

C
7.68 volumes

D
none of these

Solution

Moles of iodine= $$\cfrac {0.508}{254}$$ moles
Now, $$M_1V_1=M_2V_2$$
$$M_1\longrightarrow$$ Molarity of $$H_2O_2$$ solution
$$\therefore$$ $$V_1\longrightarrow$$ Volume of $$H_2O_2$$
$$M_2\longrightarrow$$ Molarity of iodine
$$V_2\longrightarrow$$ $$1000$$ $$ml$$

$$\therefore$$ $$M_1\times 5=\cfrac {0.508}{254}\times 1000$$
$$\therefore$$ $$M_1=0.4$$ $$M$$

The reaction can be given as,
$$H_2O_2+2I^{-}+2H^{+}\longrightarrow2H_2O+I_2$$

$$\therefore$$ n-factor for $$H_2O_2=2$$
$$\therefore$$ Normality of $$H_2O_2=2\times 0.4=0.8N$$

$$\therefore$$ Volume strength of $$H_2O_2=0.8\times 5.6$$
=$$4.48$$ volumes
Question 5
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What will be the strength of 20 vol of $$H_2O_2$$ in terms of gram per litre?

A
$$60.71g L^{-1}$$

B
$$5.6g L^{-1}$$

C
$$30.62g L^{-1}$$

D
$$17g L^{-1}$$

Solution

Concentration (in g/L) $$= \dfrac {680}{224} \times $$ volume strength
Concentration (in g/L) $$= \dfrac {680}{224} \times 20= 60.71 \ g/L$$
Question 6
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What willl be the mass of oxygen liberated by decomposition of 200 mL hydrogen peroxide solution with a strength of 34 g per litre?

A
25.5g

B
3.0g

C
3.2g

D
4.2g

Solution

Strength of the solution $$=34 $$ g/L.
1 L (or 1000 mL) of the solution contains 34 g of $$H_2O_2$$
200 mL of the solution contains $$\dfrac{34}{1000} \times 200= 6.8 $$ g of $$H_2O_2$$
$$2H_2O_2\rightarrow2H_2O+O_2$$
2 moles of $$H_2O_2$$ gives 1 mole of $$O_2$$
68g of $$H_2O_2$$ gives $$ 32$$ g of $$O_2$$
6.8g of $$H_2O_2$$ gives $$\dfrac{32}{68} \times 6.8= 3.2$$ g of $$O_2$$
Question 7
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Which of the following reactions increases production of dihydrogen from synthesis gas?

A
$$CH_4(g) + H_2O(g) \xrightarrow[Ni]{1270K} CO(g) + 3H_2(g$$)

B
$$C(s) + H_2O(g) \overset{1270K}{\rightarrow} CO(g) + H_2(g)$$

C
$$CO(g) + H_2O(g) \xrightarrow[Catalyst]{673 K} CO_2(g) + H_2(g)$$

D
None of the above

Solution

Correct Answer: Option C
Explanation:

By reacting carbon monoxide from syngas mixture with steam in the presence of iron chromate as a catalyst, the production of dihydrogen gas can be boosted.

$$CO\left( g \right)+{{H}_{2}}O\left( g \right)\xrightarrow[catalyst]{673K}C{{O}_{2}}\left( g \right)+{{H}_{2}}\left( g \right)$$

This is called the water-gas shift reaction.

Final Answer:

The correct answer is option $$(C).$$
Question 8
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Which of the following information is true for structure of $$H_2O_2$$ in solid phase?

A

B

C

D

Solution

$$H_2O_2 $$ has an open book structure with $$H-O-O$$ bond angle equal to 101.9^0 in solid phase. The $$O-O$$ single bond distance is 145.8 pm. The $$O-H$$ single bond distance is 98.8 pm.
Question 9
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HOTS
The various types of hydrides and examples of each type are given below :
Hydride type Compound
A Electron deficient i LiH
B Saline ii $$CH_{4}$$
C Electron-precise iii $$NH_{3}$$
D Interstitial iv $$B_{2}H_{6}$$
E Electron rich v CrH
Choose the correct matching from the code given below :

A
(A)- (ii), (B) - (iv), (C) - (v), (D) - (iii), (E) - (i)

B
(A)- (iv), (B) - (i), (C) - (ii), (D) - (v), (E) - (iii)

C
(A)- (iv), (B) - (iii), (C) - (v), (D) - (ii), (E) - (i)

D
(A)- (v), (B) - (iii), (C) - i(v), (D) - (ii), (E) - (i)

Solution

(A) electron deficient $$ \rightarrow B_{2}H_{6}$$
(which can accept a love pair of electron)
(B) saline (Ionic hydrides)$$ \rightarrow LiH $$
(C) Electron precise $$ \rightarrow CH_{4}$$
(which can neither accept nor give pair of $$ e^{-} 14^{th}$$ group element makes this hydride)
(D) Interstitial $$ \rightarrow CrH $$
(d block element from intersfical hydride)
(E) Electron rich $$ \rightarrow NH_{3}$$
molecule which can donate pair of electron
(A)-iv, (B)-(i), (C)-(ii), (D)-(v), (E)-(iii)
Question 10
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Match list I with list II. Choose the correct matching codes from the choices given.
List I List II
(Hydride) (Type of hydride)
A. $$BeH_{2}$$ 1. Complex
B.. $$AsH_{3}$$ 2. Lewis acid
C. $$B_{2}H_{6}$$ 3. Interstitial
D. $$LaH_{3}$$ 4. Covalent
E. $$LiAlH_{4}$$ 5. Intermediate

A
A-6, B-2, C-4, D-5, E-1

B
A-6, B-2, C-4, D-3, E-1

C
A-6, B-4, C-2, D-3, E-15

D
A-5, B-4, C-2, D-3, E-1

Solution

$$A.$$ $$BeH_2\rightarrow$$ Intermediate
$$B.$$ $$AsH_3\rightarrow$$ Covalent
$$C.$$ $$B_2H_6\rightarrow$$ Lewis acid
$$D.$$ $$LaH_3\rightarrow$$ Interstitial
$$E.$$ $$LiAlH_4\rightarrow$$ Complex