an compound that form solvated hydrogen ion in polar solvent

an acid which accepts the proton

a compound that forms hydride ion in polar solvent

an acid which donates the proton

are formed by transition metals

are stable towards water and alcohol

MCQ Questions for Class 11 Medical Chemistry Hydrogen Quiz 10

Hydrogen resembles in many of its properties with

0%
halogens
0%
alkali metals
0%
both (A) and (B)
0%
None of these

Explanation

$H_2$ like halogens gains one electron to form stable duplet.

Also, it looses one electron like alkali metals to form

$H^+$ ion.

Hence, correct option is (C).

Protonic acid is

0%
an compound that form solvated hydrogen ion in polar solvent
0%
an acid which accepts the proton
0%
a compound that forms hydride ion in polar solvent
0%
an acid which donates the proton

Triple point of water is

0%
273 K
0%
373 K
0%
203 k
0%
193 K

Maximum number of hydrogen bonding in a

$H_{2}O$ molecule is

0%
1
0%
2
0%
3
0%
4

Explanation

Maximum number of hydrogen bonding in a

$H_2O$ molecule is 4.
1675878_1757866_ans_5c191317f12545a8b437481bcb65590e.png

Systematic name of

$H_{2}O$ (oxide of hydrogen) is

0%
Water
0%
Hydrogen oxide
0%
Oxidane
0%
None of these

What is the Boiling point of water and which states of

$H_2O$ exist at the boiling temperature of the water?

0%
100 K, liquid and gas
0%
$100^{\circ} C$ , liquid and solid
0%
$100^{\circ} C$ , liquid and gas
0%
100 K, liquid and solid

Explanation

Boiling point of water is

$100^{\circ}$ and

$H_2O$ exists in Liquid and Gaseous state at it's boiling temperature.

Hence, Option "C" is the correct answer.

Ionic hydrides are usually

0%
Good electrically conductors when solid
0%
Easily reduced
0%
Good reducing agents
0%
Liquid at room temperature

Explanation

Ionic hydrides are good reducing agents because of the unstable

$H^-$ ion.

The structure of

$H_{2}O_{2}$ is

0%
planar
0%
linear
0%
spherical
0%
non-planar

Explanation

Structure of

$H_2O_2$ is non planar as shown.
1675934_1758454_ans_82a8ad8c6e024879ba7f5213a74e039d.png

The boiling point of water is exceptionally high because

0%
There is covalent bond between H and O
0%
Water molecule is linear
0%
Water molecules associate due to hydrogen bonding
0%
Water molecule is not linear

Lead pipes are not suitable for drinking water because

0%
A layer of lead dioxide is deposited over pipes
0%
Lead reacts with air to form litharge
0%
Lead reacts will water containing air to form $Pb(OH)_{2}$
0%
Lead forms basic lead carbonate

Explanation

Lead pipes are not suitable for drinking water because Lead reacts will water containing air to form

$Pb(OH)_{2}$ which is poisonous for our health.

Hence, Option "C" is the correct answer.

The oxide that gives hydrogen peroxide

$\left ( H_{2}O_{2} \right )$ on the treatment with a dilute acid

$H_{2}SO_{4}$

0%
$MnO_{2}$
0%
$PbO_{2}$
0%
$Na_{2}O_{2}$
0%
$TiO_{2}$

Explanation

The oxide that gives hydrogen peroxide

$\left ( H_{2}O_{2} \right )$ on the treatment with a dilute acid

$H_{2}SO_{4}$ is

$Na_{2}O_{2} .$

$Na_2O_{2}+H_{2}SO_{4}\rightarrow Na_{2}SO_{4}+H_{2}O_{2}$

The property of hydrogen which distinguishes it from alkali metals is

0%
Its electropositive character
0%
Its affinity for non metal
0%
Its reducing character
0%
Its non-metallic character

Hydrogen can behave as a metal

0%
At very high temperature
0%
At very low temperature
0%
At very high pressure
0%
At very low pressure

Hydrogen has a tendency to gain one electron to acquire helium configuration. In this respect it resembles

0%
Halogens
0%
Actinides
0%
Transition elements
0%
Alkali metals

The amount of

$H_{2}O_{2}$ present in 1 L of 1.5

$NH_{2}O_{2}$ solution is

0%
2.5 g
0%
25.5 g
0%
3.0 g
0%
8.0 g

Explanation

Strength of

$H_2O_2=$ Normality

$\times$ Eq. mass

$=1.5 \times 17$ (eq. mass of

$H_{2}O_{2})$

$= 25.5gL^{-1}$

Metals of groups 7, 8 and 9 do not from metallic hydrides. This is termed as

0%
hydride gap
0%
hydride shift
0%
anhydride
0%
dehydride

Reaction between the following pairs will produce

$H_2$ except

0%
Na + ethyl alcohol
0%
Fe + steam
0%
$Fe +H_2SO_4 (aq)$
0%
$Cu +HCl (aq)$

Explanation

$Cu +HCl \rightarrow No \ reaction$

Copper is less reactive than hydrogen. Therefore it is unable to displace hydrogen from acid.

$E^0_{Cu} = +0.34$ and

$E^0_H = 0.00$

Which of the following statements is correct?

0%
Elements of group 15 form electron deficient hydrides.
0%
All elements of group 14 form electron precise hydrides.
0%
Electron precise hydrides have tetrahedral geometries.
0%
Electron rich hydrides can act as Lewis acids.

Explanation

Electrons precise hydrides are those hydrides in which they have the exact number of electrons to form normal covalent bond . All elements of group 14 form such compounds (e.g.,

$CH_4$ ) which is tetrahedral in geometry.

The correct electron dot structure of a water molecule is

Only one element of

$----$ forms hydride.

0%
group 6
0%
group 7
0%
group 8
0%
group 9

Elements which show unique nature in the preparation of hydrogen are

0%
$Na, \ K, \ Li$
0%
$Mg, \ Ca, \ Ba$
0%
$Fe, \ Cu, \ Ag$
0%
$Al,\ Zn,\ Pb$

Explanation

Elements that show unique nature in the preparation of hydrogen are

$Al, \ Zn, and \ Pb$ . Aluminium, zinc and lead have unique nature. They react with acids and can even react with hot concentrated alkalis to form hydrogen and soluble salt. Salts (oxides and hydroxide) of these metals are Amphoteric.

Option D is correct.

When ice melts to form liquid water at

$0^0C$ , there is a contraction in volume. This is due to :

0%
the molecules contracting in size
0%
a partial disruption of the hydrogen bonded network of ice on melting
0%
the absorption of heat during the melting process
0%
the dissolving of air into the water during the melting process

Explanation

Ice melts to form liquid water at

$0^oC$ , there contraction in volume is due to a partial disruption of the hydrogen bonded network of ice on melting.

What is not true about ice?

0%
It has open cage like structure
0%
It has less density than water
0%
Each O atom is surrounded by 4 H atoms
0%
Each O atom has four H-bonds around it

Identify X in the reaction:

$X + \underset{(steam)}{2H_{2}O} \overset{\Delta}{\longrightarrow} XO_{2} + 2H_{2}$

0%
$C$
0%
$Si$
0%
$Ge$
0%
$Sn$

Explanation

$Sn + \underset{Steam}{2H_{2}O} \rightarrow SnO_{2} + 2H_{2}$

Tin decomposes steam to form tin dioxide and hydrogen gas.

Strength of

$10$ volume hydrogen peroxide solution means

0%
$30.35\ g\ L^{-1}$
0%
$17\ g\ L^{-1}$
0%
$34\ g\ L^{-1}$
0%
$68\ g\ L^{-1}$

Explanation

$2H_2O_2\to 2H_2O+O_2$

$22.4\ O_2$ is given by

$68\ g$ of

$H_2O_2$

$10\ L\ O_2$ is given by

$\dfrac {68}{22.4}\times 10=30.35\ g\ L^{-1}$

Which is not a property of water?

0%
It is a colourless and tasteless liquid.
0%
There is no hydrogen bonding in solid state of water.
0%
It is an excellent solvent for transportation of ions in plants and animals.
0%
Frozen water is lighter than liquid water.

Heavy water

$(D_2O)$ freezes at

0%
$-3.8^o\ C$
0%
$3.8^o\ C$
0%
$0^o\ C$
0%
$38^o\ C$

Explanation

ANSWER is

$option$ (B)

Heavy water freezes at

$3.8^o\ C$ .

Water contains 88.89% of A by weight is decomposed into its elements B and oxygen. Pure C is a poor conductor of electricity and hence it is acidified using dilute sulphuric acid. When electric current is passed through acidulated water, using platinum electrones, hydrogen gas is evolved at cathode and oxygen gas is evolved at D.

What is D?

0%
Anode
0%
Cathode
0%
Electrolyte
0%
Catalyst

Explanation

Percentage of oxygen in water

$\displaystyle = \frac{weight\ of\ oxygen\ in\ water}{weight\ of\ water} \times 100$

$\displaystyle = \frac{16}{18} \times 100$

$=$ 88.89%

Thus, water contains 88.89% of oxygen (A)

On decomposition of water we get hydrogen and oxygen.

So, B is hydrogen

Pure water is a poor conductor of electricity and when acidified by adding sulphuric acid it can conduct electricity hence, C is water.

During electrolysis of water, hydrogen gas is evolved at cathode and oxygen gas is evolved at anode.

Since at D oxygen gas is evolved D is anode.

Pure hydrogen is obtained by carrying electrolysis of

0%
Water containing $H_{2}SO_{4}$
0%
Water containing $NaOH$
0%
$Ba\left ( OH \right )_{2}$ solution
0%
$KOH$ solution

Explanation

Pure hydrogen is obtained by the electrolysis of

$Ba(OH)_2$ solution in a U-tube using nickel electrode. The gas is liberated at the cathode and is passed over heated platinum gauze to remove oxygen if present as impurity.

Limiting compositions of f-block hydrides are

0%
$MH_{2}$ and $MH_{3}$
0%
$MH_{3}$ and $MH_{5}$
0%
$MH_{2}$ and $MH_{8}$
0%
$MH_{2}$ and $MH_{6}$

Explanation

Limiting compositions of f-block hydrides are

$MH_2$ and

$MH_3$ .

Water contains 88.89% of A by weight is decomposed into its elements B and oxygen. Pure C is a poor conductor of electricity and hence it is acidified using dilute sulphuric acid. When an electric current is passed through acidulated water, using platinum electrons, hydrogen gas is evolved at cathode and oxygen gas is evolved at D.

What is A?

0%
Hydrogen
0%
Oxygen
0%
Water
0%
Sulphuric acid

Explanation

Percentage of oxygen in water

$\displaystyle = \frac{weight\ of\ oxygen\ in\ water}{weight\ of\ water} \times 100$

$\displaystyle = \frac{16}{18} \times 100$

$=$ 88.89%

Thus, water contains 88.89% of oxygen (A).

Saline hydrides:

0%
are dehydrating agents
0%
are formed by transition metals
0%
are very poor reducing agents
0%
are stable towards water and alcohol

Which is true statement about

$H_{2}O$ ?

0%
Hardness can be removed by passing through ion exchange resin
0%
Its pressure can be detected by anhydrous $CuSO_{4}$
0%
It is amphoprotic
0%
All of the above

Explanation

Its hardness can be removed by passing through ion exchange resin, its pressure can be detected by anhydrous

$CuSO_4$ .

$H_2O$ is amphoprotic as it can be oxidised as well as reduced or it can act as acid as well as base.

What is 'B'?

0%
Oxygen
0%
Hydrogen
0%
Water
0%
Sulphuric acid

Explanation

Percentage of oxygen in water

$\displaystyle = \frac{weight\ of\ oxygen\ in\ water}{weight\ of\ water} \times 100$

$\displaystyle = \frac{16}{18} \times 100$

$=$ 88.89%

Thus, water contains 88.89% of oxygen (A)

On decomposition of water we get hydrogen and oxygen.

So, B is hydrogen

Which of the following statement(s) is/are correct about the water-gas shift reaction which is used in the commercial preparation of hydrogen gas?

0%
It involves the selective oxidation of $CO$ of water gas to $CO_{2}$ with the help of steam in the presence of iron chromate as catalyst
0%
It involves the separation of $CO$ and $H_{2}$ by solvent extraction method
0%
It involves the absorption of $CO$ in aqueous copper $(I)$ chloride solution
0%
It involves the removal of $H_{2}$ through occlusion with $Pd$

Explanation

$CO_{(g)} + H_{2(g)} + H_2 O_{(g)} \xrightarrow [673\: K] {FeCrO_4} 2 H_{2(g)} + CO_{2(g) }$

Water-gas shift reaction involves the selective oxidation of

$CO$ of water gas to

$CO_2$ with the help of steam, in the presence of iron chromate as catalyst.

Half litre each of three samples of

$H_{2}O_{2}$ labelled

$10$ volume,

$15$ volume and

$20$ volume are mixed and then diluted with equal volume of water. The relative strength of resultant

$H_{2}O_{2}$ solution is:

0%
$2.5$
0%
$5.0$
0%
$7.5$
0%
$10$

Explanation

Volume strength of

$H_{ 2 }{ O }_{ 2 } = 5.6\times \text{N}$

$(\text{N} =$ Normality

$)$

Normality of

$10$ vol of

$H_{ 2 }{ O }_{ 2 } = \dfrac{10}{5.6}\text{N}$

Normality of

$15$ vol of

$H_{ 2 }{ O }_{ 2 } = \dfrac{15}{5.6} \text{N}$

Normality of

$20$ vol of

$H_{ 2 }{ O }_{ 2 } = \dfrac{20}{5.6} \text{N}$

Let

$500$ ml of each is mixed, then the total volume of mix

$= 1500$ ml.

Also, this is diluted to double its volume, i.e.,

$= 3000$ ml

$\text{N}\times 3000$ =

$\dfrac { 10 }{ 5.6 } \times 500+\dfrac { 15 }{ 5.6 } \times 500+\dfrac { 20 }{ 5.6 } \times 500$

Normality of

$H_{ 2 }{ O }_{ 2 }$ mix

$= 1.339\:\text{N}$

Vol. strength of

$H_{ 2 }{ O }_{ 2 } = 1.339 \times 5.6 = 7.5$ .

$10$ ml of

$H_2O_2$ solution requires

$25$ ml of

$\dfrac{\text{N}}{20}\: KMnO_{4}$ for complete oxidation in acidic medium, then the mass (in g) of available

$O_{2}$ per litre present in the solution is:

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

The oxidation reaction is as follows:

$2KMn{ O }_{ 4 } + 5{ H }_{ 2 }{ O }_{ 2 }+ 4{ H }_{ 2 }{ SO }_{ 4 }\rightarrow 2{ KHSO }_{ 4 }+ 2MnS{ O }_{ 4 }+ 8{ H }_{ 2 }O+ 5{ O }_{ 2 }$

Meq. of

${ O }_{ 2 } =$ Meq. of

${ H }_{ 2 }{ O }_{ 2 } =$ Meq. of

$KMn{ O }_{ 4 }$

It is given that

$10$ ml of

$H_2O_2$ solution when treated with

$25$ ml of

$\dfrac {N}{20}\ KMnO_4$ for complete oxidation.

$\text{N} \times 10 = 25 \times \dfrac{1}{20}$

Normality of

${ H }_{ 2 }{ O }_{ 2 } = 0.125\:\text{N} =$ Normality of

${ O }_{ 2 }$

Weight of

${ O }_{ 2 }= 0.125 \times 32 = 4$ g per litre.

$25$ ml of

$H_{2}O_{2}$ solution were added to excess of acidified

$KI$ solution. The iodine so liberated required

$20$ ml of

$0.1\: \text{N}$

$Na_{2}S_{2}O_{3}$ for titration. Calculate strength of

$H_2O_2$ in terms of normality and percentage?

0%
$0.04\:\text{N}, 0.136\%$
0%
$0.08\:\text{N}, 0.136\%$
0%
$0.04\:\text{N}, 0.163\%$
0%
$0.02\:\text{N}, 0.163\%$

Explanation

$I_{2}+2Na_{2}S_{2}O_{3}^-\rightarrow Na_{2}S_{4}O_{6}+2NaI$

Milliequivalent of

$H_2O_2 =$ Milliequivalent of

$I_2 =$ Milliequivalent of

$Na_2S_2O_3$

$\dfrac { w }{ 34/2 } \times 1000 = 20 \times 0.1$

${ \therefore W }_{ H_{ 2 }{ O }_{ 2 } } = 0.034$ gm

$/ 25$ ml

${ N }_{ H_{ 2 }{ O }_{ 2 } } = \dfrac { 0.034 }{ 34/2 } \times \dfrac { 1000 }{ 25 } = 0.08$

Volume strength

$= 5.6\times0.08 = 0.448$

$\%$ strength

$= \dfrac { 17 }{ 56 } \times 5.6 \times 0.08 = 0.136%$ .

Peroxoborate in aqueous solution provides:

0%
$[B(OH)_4]^-+H_2O_2$
0%
$Na_2B_4O_8\cdot 10H_2O$
0%
$HBO_2$
0%
$Na[B_3O_3(OH)_4]$

Explanation

Hence, peroxoborate,

$B_2O_4(OH)_4$ in aqueous solution provides peroxide,

$H_2O_2$ and

${B(OH)_4}^-$ .

An experimental set up is shown in the figure. Identify the metal X and gas Y respectively in the figure.

0%
Zinc, $\displaystyle H_{2}$ gas
0%
Copper, $\displaystyle CO_{2}$ gas
0%
Silver, $\displaystyle CO_{2}$ gas
0%
Platinum, $\displaystyle H_{2}$ gas

Explanation

Zinc reacts with dil. HCl to give hydrogen gas which is collected above water as it is the lightest gas while other metals are unreactive towards dil. HCl as they are inert present below the hydrogen in the reactivity series.

$Zn(s) +HCl(dil.)\rightarrow ZnCl_2(aq) +H_2(g)$

Option A is correct.

By which process

$H_{2}O_{2}$ does not form?

0%
0%
0%
0%
By $(A)$ and $(C)$ process

Explanation

Hydrogen peroxide is prepared by Merck's method in which sodium peroxide reacts with dilute acids.

$Na_2O_2+H_2SO_4 \rightarrow Na_2SO_4 + H_2O_2$ .

While

$KO_2$ and

$PbO_2$ does not form hydrogen peroxide on reacting with dilute acids.

Option D is correct.

A teacher one day pointed out to his students the peculiar fact that water is a unique liquid that freezes exactly at

$0^oC$ and boils exactly at

$100^oC$ . He asked the students to find the correct statement based on this fact.

Which of the statements is true regarding the above fact?

0%
Water dissolves anything however sparingly the dissolution may be.
0%
Water is a polar molecule.
0%
Boiling and freezing temperatures of water were used to define a temperature scale.
0%
Liquid water is denser than ice.

When a substance A reacts with water, it produces a combustible gas B and a solution of substance C in water. D reacts with this solution of C and produces the same gas B on warming. D can also produce gas B on reaction with dilute $\displaystyle H_{2}SO_{4}$ . A imparts a deep golden yellow colour to smokeless flame. A, B, C and D respectively are:

0%
$Na \displaystyle, H_{2}, NaOH, Zn$
0%
$K, \displaystyle H_{2}, KOH, Al$
0%
$Ca, \displaystyle H_{2}, \displaystyle Ca\left ( OH \right )_{2}, Sn$
0%
$CaC_{2},\displaystyle C_{2}H_{2},\displaystyle Ca\left ( OH \right )_{2}, Fe$

Explanation

Reactions:

$2Na+2H_2O\rightarrow 2NaOH+H_2$

A C B

$Na$ gives a golden colour flame.

$Zn+2NaOH(dil.)\rightarrow Na_2ZnO_2+H_2\uparrow$

D C B

Which of the following produces hydrolith with dihydrogen?

0%
Mg
0%
Al
0%
Cu
0%
Ca

Explanation

Calcium reacts with dihydrogen to produce hydrolith which is calcium hydride.

$Ca+H_2 \xrightarrow {\Delta} CaH_2$

$\displaystyle { H }_{ 2 }O$ has a higher boiling point than HF. Which among the following is the right reason ?

0%
$\displaystyle { H }_{ 2 }O$ is more polar than HF
0%
$\displaystyle { H }_{ 2 }O$ can form more hydrogen bonds
0%
$\displaystyle { H }_{ 2 }O$ has a higher molecular weight
0%
$\displaystyle { H }_{ 2 }O$ has more atoms
0%
$\displaystyle { H }_{ 2 }O$ does not have a higher boiling point than HF

Explanation

B is right option and A is wrong.
As

$H_2O$ is less polar than the

$HF$ , but

$H_2O$ can form more hydrogen bonds than

$HF$ .
A water molecule has twice as many hydrogen atoms as does HF and, therefore, water can theoretically form more hydrogen bonds than hydrogen fluoride. More hydrogen bonds mean more intermolecular attractions, and more intermolecular attractions mean a higher boiling point.
Hence , we can say that

$H_2O$ has higher extent of

$H-bonding$ than the

$HF$ .

What is the condensation point of water?

0%
0 degrees Celsius
0%
0 degrees Fahrenheit
0%
100 degrees Fahrenheit
0%
212 degrees Fahrenheit

Explanation

Condensation point is a temperature at which gas changes to liquid. Generally, boiling point is same as condensation point. For water boiling point is 100

$^{\circ}C$ , -212

$^{\circ}F$ . So, condensation point is also 212

$^{\circ}F$ .

Boiling water reacts with _____________ gives their respective hydroxides and ammonia gas.

0%
magnesium nitride
0%
calcium nitride
0%
aluminium nitride
0%
all of the above

Boiling water reacts with

$C_6H_5N^+_2Cl^-$ to give

0%
Aniline
0%
Benzylamine
0%
Phenol
0%
Benzaldehyde

What is the volume of "20 volume

$H_2O_2$ " required to get

$5000\ cm^3$ of oxygen at S.T.P?

0%
$250\ cm^3$
0%
$50\ cm^3$
0%
$100\ cm^3$
0%
$125\ cm^3$

Explanation

$\displaystyle 5000\ cm^3$ of oxygen corresponds to 5 L.

20 volume

$\displaystyle H_2O_2$ means that decomposition of 1 L of

$\displaystyle H_2O_2$ gives 20 L of oxygen at STP.

5 L of oxygen at S.T.P will be obtained from decomposition of

$\displaystyle \dfrac {5}{20} \times 1 = 0.250 \: L \: H_2O_2$ .

This corresponds to

$\displaystyle 250 \: cm^3$ of 20 volume

$\displaystyle H_2O_2$ .

Choose the correct statement about the industrial preparation of hydrogen from water gas

$(CO+H_2)$ .

0%
$H_2$ is removed through occlussion with $Pd$
0%
$CO$ and $H_2$ are fractionally separated using differences in their densities
0%
$CO$ is removed by absorption in $aq. Cu_2Cl_2$ solution
0%
$CO$ is oxidised to $CO_2$ with steam in the presence of a catalyst followed by absorption of $CO_2$ in alkali

Explanation

$CO$ is first oxidised to

$CO_2$ with steam in the presence of a catalyst followed by absorption of

$CO_2$ in alkali.

CBSE Questions for Class 11 Medical Chemistry Hydrogen Quiz 10 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Hydrogen Quiz 10 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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