Hydrogen Test

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Hydrogen Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

$$3.4$$ g of $$H_{2}O_{2}$$ decomposes, the weight of oxygen liberated from it is:

A
$$1.6$$ g

B
$$2.24$$ g

C
$$1.16$$ g

D
$$3.2$$ g

Solution

$$2{ H }_{ 2 }{ O }_{ 2 } \rightarrow 2{ H }_{ 2 }O +{ O }_{ 2 }$$.
$$68$$ g of $${ H }_{ 2 }{ O }_{ 2 }$$ produce --- $$32$$ g of $${ O }_{ 2 }$$
$$3.4$$ g of $${ H }_{ 2 }{ O }_{ 2 }$$ produce --- $$\dfrac { 32 }{ 68 } \times 3.4 = 1.6$$ g of $${ O }_{ 2 }$$.
Question 2
1 / -0

The volume of perhydrol which on decomposition gives 1.5 lit of $$O_{2}$$ gas at STP is:

A
$$25$$ ml

B
$$15$$ ml

C
$$10$$ ml

D
$$20$$ ml

Solution

As, it is a perhydrol sample, $$1$$ ml of $$H_{2}O_{2}$$ would give $$100$$ ml of $$O_{2}$$ at room temperature. Thus, $$1.5$$ L of $$O_{2}$$ would be given by $$15$$ ml of $$H_{2}O_{2}$$.
Question 3
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The volume strength of $$15\%$$ (w/v) solution of $$H_{2}O_{2}$$ is :

A
$$50$$

B
$$4.94$$

C
$$494$$

D
$$0.494$$

Solution

The solution is $$15\% (w/v)$$. It means $$100 \ ml$$ of solution contains $$15$$ g of hydrogen peroxide.
The concentration in g/L is $$\dfrac {15}{100} \times 1000 =150 g/L$$.
The relationship between the concentration in g/L and the volume strength $$\text{x}$$ is
Conc. in g/L $$= \dfrac {680\times \text{(x)}}{224} g/L$$

Thus, $$150 = \dfrac {680\times \text{(x)}}{224}$$
$$\text{x}=50$$. Thus, the volume strength of solution is $$50$$.
Question 4
1 / -0

The volume of $$20\:V-H_{2}O_{2}$$ solution, required to prepare $$5$$ litres of oxygen at STP is:

A
$$125$$ ml

B
$$250$$ ml

C
$$100$$ ml

D
$$50$$ ml

Solution

$$20$$ ml of $${ O }_{ 2 }$$ is prepared by $$1$$ ml of $${ H }_{ 2 }{ O }_{ 2 }$$ at STP.

$$1$$ ml of $${ O }_{ 2 }$$ is prepared by $$= \dfrac{1}{20}$$ ml of $${ H }_{ 2 }{ O }_{ 2 }$$.

$$5000$$ ml of $${ O }_{ 2 }$$ is prepared by = $$\dfrac { 1 }{ 20 } \times 5000 = 250$$ ml of $${ H }_{ 2 }{ O }_{ 2 }$$.

Hence, option B is correct.
Question 5
1 / -0

The volume (in ml) of oxygen liberated at STP when $$100$$ ml of $$15\:V$$ $$H_{2}O_{2}$$ decomposes is:

A
$$150$$

B
$$300$$

C
$$3000$$

D
$$1500$$

Solution

$$1$$ ml $$H_2O_2$$ of $$15$$ vol liberates $$15$$ ml of $$O_2$$ at STP.
Hence, $$100$$ ml $$H_2O_2$$ of $$15$$ vol liberates $$15\times100$$ ml of $$O_2$$ at STP or $$1500$$ ml of $$O_2$$ at STP.
Hence, option D is the correct answer.
Question 6
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The volume strength of $$3N\:H_{2}O_{2}$$ solution is:

A
$$11.2\:\text{V}$$

B
$$8.4\:\text{V}$$

C
$$22.4\:\text{V}$$

D
$$16.8\:\text{V}$$

Solution

$$3\:N$$ means $$3$$ g eqwt. in $$1$$ L solution.

$$N=\dfrac{\text{weight}}{\text{eq. wt.}}$$.

$$($$eq. wt. for hydrogen peroxide$$ = 17$$ g$$)$$

Therefore, $$3 = \dfrac{\text{wt}}{17}$$ or $$\text{wt} = 3 \times17 = 51$$ g.

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

Therefore, $$51$$ g of hydrogen peroxide gives $$22.4 \times \dfrac{51}{68} = 16.8$$ L.

Thus, volume strength is $$16.8$$ L, because volume strength is the volume of oxygen at STP given by one volume sample of hydrogen peroxide on heating.
Question 7
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The $$\%$$ by weight of hydrogen in $$H_{2}O_{2}$$ is :

A
$$50$$

B
$$25$$

C
$$6.25$$

D
$$5.88$$

Solution

The molecular weight of $$H_2O_2$$ is $$34$$ g.
It has $$2$$ g of $$H_2$$ in it.
Therefore, percentage weight of $$H_2$$ $$=$$ $$\dfrac{2}{34}\times100 = 5.88\%$$
Hence, the percentage weight of hydrogen in $$H_2O_2$$ is $$5.88\%$$.
Question 8
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$$H_{2}O_{2}$$ turns blackened lead paintings to white colour. In this reaction it oxidizes $$PbS$$ to $$PbSO_{4}$$. The number of moles of $$H_{2}O_{2}$$ needed to oxidize $$0.1$$ moles of $$PbS$$ is:

A
1 mole

B
0.1 mole

C
0.5 mole

D
0.4 mole

Solution

The balanced equation is:

$$PbS+4H_{2}O_{2}\rightarrow PbSO_{4}+4H_{2}O$$.

The stiochiometric coefficient of $$PbS$$ is $$1 $$ while that of $$H_{2}O_{2}$$ is $$4$$ in the balanced reaction.

So, 0.1 mol will be oxidised by 0.4 mol of $$H_2O_2$$
Question 9
1 / -0

The co-ordination number of $$Na$$ in solid $$NaH$$ is:

A
$$4$$

B
$$6$$

C
$$8$$

D
$$12$$

Solution

Coordination number of $$Na$$ in $$NaH$$ is $$6$$ as its shape is octahedral.
Question 10
1 / -0

The reaction between $$H_2O_2$$ and $$KMnO_4$$ is $$2KMnO_4+3H_2SO_4+5H_2O_2 \rightarrow K_2SO_4+2MnSO_4+8H_2O+5O_2$$.
In a reaction excess of $$H_2O_2$$ is added to 0.1 mole of acidified $$KMnO_4$$ solution. Then the STP volume of $$O_2$$ liberated is:

A
5.6 L

B
6.6 L

C
11.2 L

D
22.4 L

Solution

$$0.1$$ mole of $$KMnO_{4}$$ produces $$0.25$$ moles of $$O_{2}$$ which has a volume of 22.4 x 0.25 litres at STP.

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Hydrogen Test - 20

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Hydrogen Test - 20

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Question 1

$$1.6$$ g

$$2.24$$ g

$$1.16$$ g

$$3.2$$ g

Solution

Question 2

$$25$$ ml

$$15$$ ml

$$10$$ ml

$$20$$ ml

Solution

Question 3

$$50$$

$$4.94$$

$$494$$

$$0.494$$

Solution

Question 4

$$125$$ ml

$$250$$ ml

$$100$$ ml

$$50$$ ml

Solution

Question 5

$$150$$

$$300$$

$$3000$$

$$1500$$

Solution

Question 6

$$11.2\:\text{V}$$

$$8.4\:\text{V}$$

$$22.4\:\text{V}$$

$$16.8\:\text{V}$$

Solution

Question 7

$$50$$

$$25$$

$$6.25$$

$$5.88$$

Solution

Question 8

1 mole

0.1 mole

0.5 mole

0.4 mole

Solution

Question 9

$$4$$

$$6$$

$$8$$

$$12$$

Solution

Question 10

5.6 L

6.6 L

11.2 L

22.4 L

Solution

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