Hydrogen Test

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Hydrogen Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

The concentration of $$H_{2}O_{2}$$ in a solution containing 34 gm in 500ml is given in List I and their corresponding values are given in List II. Then match List I with List II.

List I List II
A) Molarity
1) 68
B) Normality
2) 22.4
C) % w/v
3) 4
D) Volume strength
4) 2

5) 10

A
(A)-5, (B)-3, (C)-1, (D)-2

B
(A)-4, (B)-3, (C)-1, (D)-2

C
(A)-3, (B)-4, (C)-1, (D)-2

D
(A)-1, (B)-2, (C)-3, (D)-4

Solution

(A) The molar mass of hydrogen peroxide is 34 g/mol.
The molarity of 34 g of hydrogen peroxide in 500 ml is $$\displaystyle \frac {34g}{34g/mol} \times \frac {1000mL/L}{500mL} = 2$$ M.

(B) The normality of hydrogen peroxide is twice its molarity. It is $$\displaystyle 2 \times 2 = 4$$ N.

(C) Percent w/v of hydrogen peroxide solution is $$\displaystyle \frac {34 g \times 1000mL/L}{500mL} = 68$$ %

(D) Volume strength $$\displaystyle = 11.2 \times$$ molarity $$\displaystyle = 11.2 \times 2 = 22.4$$
Question 2
1 / -0

Strength of 2N $$H_{2}O_{2}$$ solution is approximately:

A
10 volumes

B
44.8 volumes

C
22.4 volumes

D
11.2 volumes

Solution

volume strength is referred to the volume of oxygen gas liberated from 1 unit volume of $$2H_2O_2$$ solution.
According to the reaction
$$2H_2O_2 \rightarrow 2H_2O+O_2$$
$$1$$ mol of $$H_2O_2$$ produces $$0.5$$ mol or $$11.2\ L$$ of $$O_2$$.
The strength of the compound $$=11.2 \times$$molarity of the solution
We have, $$2N$$ $$H_{2}O_{2}$$ $$=1M$$ $$H_{2}O_{2}$$
$$\therefore$$ strength of $$2N$$ $$H_{2}O_{2}=$$11.2 volumes.
Question 3
1 / -0

The number of electrons involved in the manufacture of 1 mole of $$H_{2}O_{2}$$ from 50% $$H_{2}SO_{4}$$:

A
2 moles

B
3 moles

C
1 mole

D
4 moles

Solution

The equations are:

$$2H_{2}SO_{4}\rightarrow 2HSO_{4}^{-}+2H^{+}$$

$$2HSO_{4}^{-}\rightarrow H_{2}S_{2}O_{8}+2e^{-}$$ At anode

$$2H^{+}+2e^{-}\rightarrow H_{2}$$ At cathode

$$H_{2}S_{2}O_{8}+2H_{2}O\rightarrow 2H_{2}SO_{4}+H_{2}O_{2}$$

Hence, two moles of electrons are transferred in the formation of 1 mole of $$H_{2}O_{2}$$.
Question 4
1 / -0

The weight - volume percentage (w/v) of $$10\ V$$ $$H_{2}O_{2}$$ solution is:

A
$$6$$%

B
$$2$$%

C
$$11$$%

D
$$3$$%

Solution

The volume strength of $$H_2O_2$$ can be converted into the weight - volume percentage (w/v) by the following expression:

The weight - volume percentage (w/v) $$= \dfrac {68x}{224}$$ %

Where $$x$$ represents volume strength which in this example is $$10 V$$.

The weight - volume percentage (w/v) $$= \dfrac {68 \times 10}{224}= 3$$ %
Question 5
1 / -0

Which ionic hydride is stable upto its melting point?

A
$$NaH$$

B
$$CaH_{2}$$

C
$$LiH$$

D
$$BaH_{2}$$

Solution

All ionic hydrides of group $$I$$ and group $$II$$ decompose before their melting point except $$LiH$$.
$$LiH$$ is stable upto its melting point.
The ionic character of the bonds in hydrides increases from $$LiH$$ to $$CsH$$ and their stability decreases in the same order.
Question 6
1 / -0

Weight of $$H_{2}O_{2}$$ present in $$560$$ ml of $$20$$ volume $$H_{2}O_{2}$$ solution is approximately:

A
69 g

B
34 g

C
32 g

D
3.4 g

Solution

Volume strength is referred to the volume of oxygen gas liberated from 1 mL volume of $$2H_2O_2$$ solution.

According to the reaction
$$2H_2O_2 \rightarrow 2H_2O+O_2$$

$$1\ mol$$ ($$34\ g$$) of $$H_2O_2$$ produces $$0.5$$ mol or $$11200\ mL$$ of $$O_2$$.

$$20\ mL$$ oxygen will be obtained from :

$$\frac {34}{11200} \times 20=0.061\ g$$

Therefore, $$0.061g$$ of $$H_2O_2$$ in $$1\ mL$$ gives 20 volume $$2H_2O_2$$

Therefore in $$560\ mL$$ solution, amount of $$H_2O_2$$ present = $$560 \times 0.061=34.16g$$

Option B is correct.
Question 7
1 / -0

The volume strength of $$1N$$ $$H_{2}O_{2}$$ solution is:

A
$$11.2\: V$$

B
$$22.4\: V$$

C
$$1\: V$$

D
$$5.6\: V$$

Solution

$$1 N$$ means $$1 g$$ equivalent weight in $$1 L$$ solution.

$$N= \dfrac{weight}{eq.wt}$$ (eq. wt for hydrogen peroxide $$=17g$$)

Therefore, $$1 = \dfrac{wt}{17}$$ or $$wt = 1\times 17 = 17 g$$

Now, from the reaction, we know that $$68 g$$ of $$H_2O_2$$ gives $$22.4 L$$ of Oxygen.

Therefore, $$17 g$$ of hydrogen peroxide gives $$\dfrac{22.4 \times 17}{68} = 5.6 L$$.

Thus, the volume strength is $$5.6L$$ because volume strength is the volume of oxygen at STP given by one volume sample of hydrogen peroxide on heating.
Question 8
1 / -0

$$50$$ ml of a certain sample of $$H_{2}O_{2}$$ gives $$1200$$ ml of $$O_{2}$$ at STP. The volume strength of $$H_{2}O_{2}$$ is:

A
22.4 V

B
20 V

C
24 V

D
11.2 V

Solution

It is given that $$50$$ ml of a certain sample of $$H_2O_2$$ gives $$1200$$ ml of $$O_2$$ at STP.

Volume strength is the volume of oxygen at STP given by one volume sample of hydrogen peroxide on heating.

If $$50$$ ml of $$H_2O_2$$ gives $$1200$$ ml of $$O_2$$, then volume strength will be $$= \dfrac {1200}{50}= 24\ V$$

Hence, the volume strength of $$H_2O_2$$ is $$24\ V$$.
Question 9
1 / -0

Elements of which of the following groups do not form hydrides?

A
$$1,2,3$$

B
$$3,4,5$$

C
$$7,8,9$$

D
$$6,7,8$$

Solution

The region of the periodic table from the group $$7$$ to $$9$$ is referred to as the hydride gap as they do not form hydrides.

Examples of such elements are Mn, Fe, Co, Ru etc. These elements do not form hydrides on account of low affinity for hydrogen in their normal oxidation states.

Hence, option C is correct.
Question 10
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Statement-1: $$CaH_{2}$$ is used as main source to transport gas as 1 pound of it gives 16.5 cubic feet $$H_{2}$$.
Statement-2: It is difficult to transport $$H_{2}$$ gas.

A

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

B
Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1.

C

Statement-1 is true, Statement-2 is false.

D
Statement-1 is false, Statement-2 is true.

Solution

Statement-1: $$CaH_{2}$$ (also called hydrolith is used as main source to transport gas as 1 pound of it gives 16.5 cubic feet $$H_{2}$$.
Statement-2: It is difficult to transport $$H_{2}$$ gas due to safety considerations as hydrogen is highly inflammable..
Hence, statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

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Re-Attempt Weekly Quiz Competition

A) Molarity	1) 68
B) Normality	2) 22.4
C) % w/v	3) 4
D) Volume strength	4) 2
	5) 10

Hydrogen Test - 21

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Hydrogen Test - 21

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Question 1

(A)-5, (B)-3, (C)-1, (D)-2

(A)-4, (B)-3, (C)-1, (D)-2

(A)-3, (B)-4, (C)-1, (D)-2

(A)-1, (B)-2, (C)-3, (D)-4

Solution

Question 2

10 volumes

44.8 volumes

22.4 volumes

11.2 volumes

Solution

Question 3

2 moles

3 moles

1 mole

4 moles

Solution

Question 4

$$6$$%

$$2$$%

$$11$$%

$$3$$%

Solution

Question 5

$$NaH$$

$$CaH_{2}$$

$$LiH$$

$$BaH_{2}$$

Solution

Question 6

69 g

34 g

32 g

3.4 g

Solution

Question 7

$$11.2\: V$$

$$22.4\: V$$

$$1\: V$$

$$5.6\: V$$

Solution

Question 8

22.4 V

20 V

24 V

11.2 V

Solution

Question 9

$$1,2,3$$

$$3,4,5$$

$$7,8,9$$

$$6,7,8$$

Solution

Question 10

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1.

Statement-1 is true, Statement-2 is false.

Statement-1 is false, Statement-2 is true.

Solution

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