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120551.

Plants in a garden make use of 22.4 liters of carbon-dioxide at STP. The numbers of molecules of carbon-dioxide utilized by the plant is (a) 6.023 x 1023 (b) 22.4 molecules (c) 6.023 x 1024 (d) 11.2 moles of molecules

Answer»

(a) 6.023 x 1023

120552.

A student writes “preparation of CO2 in the laboratory in his book. From the point of view of chemistry what is the mistake made in writing ?

Answer»

From the point of view of the mole concept he/she should have written how many mole of CO2 he/she is going to prepare.

120553.

What is valency?

Answer»

Every element reacts with other element according to its combining capacity, which we call as its valency.

120554.

Name a greenhouse gas with molar mass 44 g mol–1 and is known to extinguish fire.[Given, atomic masses of C = 12 u, O = 16 u, NA = 6.022 × 1023 per mole

Answer»

Carbon dioxide

Carbon dioxide is a greenhouse gas and is used to 

extinguish fire. 

The molar mass of CO2 is: 

Atomic mass of C + 2 × (Atomic mass of oxygen) 

⇒ 12 + 2×16 

⇒ 44g/mol

120555.

Calculate the number of moles in 360 g of this gas sample. [Given, atomic masses of C = 12 u, O = 16 u, NA = 6.022 × 1023 per mole]

Answer»

Given:

Mass of gas = 360g 

Molar mass = 44g/mol 

To calculate the number of moles, we apply the formula: 

Number of moles = \(\frac{Mass\,of\,CO_2}{Molar\,mass\,of\,CO_2}\)

⇒ Number of moles \(\frac{360g}{44g/mol}\)

⇒ Number of moles = 8.1mol

Thus, the number of moles in 360g of the sample is 8.1

120556.

Define ‘molecular mass’ of a substance.

Answer»

The molecular mass of a substance is the relative mass of its molecule as compared with the mass of a Carbon-12 atom taken as 12 units.

120557.

Mass of one molecule of a substance is 5.32 × 10-23 g. What is its molecular mass ? 

Answer»

Mass of 1 molecule of substance 

= 5.32 × 10-23

Mass of 6.022 × 1023 molecules of substance 

= 5.32 × 10-23 × 6.022 × 1023 

= 32 g

120558.

Observe the following flow chart and answer questions.(i) Based on the above flow chart, generalised statement is P : In any substance the no. of particles in one mole is constant. Q : Irrespective of quantity of a matter, the no. of atoms or molecules in it is constant. R : There is no relation between the mole and the substance. A) Q only B) P and Q C) R only D) P only(ii) Thi the above flow chart, the number 6.022 × 1023 is A) Avogadro number (N) B) Mass number (A) C) Atomic number (Z)D) Atomic mass unit (amu)

Answer»

(i) A) Q only

(ii) A) Avogadro number (N)

120559.

What is the Avogadro number ?

Answer»

The number of particles present in one mole of any substance is the Avogadro number (Na) which is equal to 6.022 × 1023 .

120560.

In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water, and 8.2 g of sodium ethanoate. Show that these observations are in agreement with law of conservation of mass.

Answer»

Sodium carbonate + Ethanoic acid → Sodium ethanoate + Carbon dioxide + Water 

The total mass of reactants = 5.3 + 6 = 11.3 g 

The total mass of products = 2.2 + 0.9 + 8.2 = 11.3 g 

∴ Total mass of reactants = Total mass of products 

So these observations are in agreement with law of conservation of mass.

120561.

In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.Sodium carbonate + ethanoic acid ➝ sodium ethanoate + carbon dioxide + water

Answer»

Sodium carbonate + ethanoic acid ➝ sodium ethanoate + carbon dioxide + water

5.3 + 6 ➝ 8.2 + 2.2 + 0.9

= 11.3 g → 11.3 g

Weight of reactants is equal to weight of products. This observation is in agreement with the law of conservation of mass.

120562.

Give definition of Plaster of Paris?

Answer»

Plaster of Paris: On heating gypsum at 373 K it loses water molecules and becomes Calcium Sulphate hemihydrate called Plaster of Paris.

120563.

Match the following:1.gram molecular(a) triatomic oxygen volume2.ozone(b) Three electrons3.valence electrons(c) diatomic oxygen4.modem standard atomic mass(d) electrons of the outer most shell(e) 22.4 litres(f) carbon 6C14 isotope.

Answer»

1. (f) 

2. (a) 

3. (d) 

4. (e)

120564.

Fill in the blanks with suitable words:Gram molecular volume of a gas at STP ………….. liter.

Answer»

Gram molecular volume of a gas at STP 22.4 liter.

120565.

ElementWeight percentageNatural sampleSynthetic sampleCopper51.3551.35Carbon9.749.74Oxygen38.9138.91From the table particularly we can prove that A) Law of conservation of mass B) Law of constant proportions C) Law of conservation of energy D) Above all

Answer»

B) Law of constant proportions

120566.

Can you write the formula of carbon dioxide and carbon monoxide? Try to write formula for them as we have done in case of water molecule.

Answer»

Carbon dioxide : 

The elements present are carbon and oxygen. One atom of carbon and one atom of oxygen are present in a molecule of carbon monoxide. Hence the formula of carbon monoxide is CO.

Carbon dioxide : 

The elements present are carbon and oxygen. One atom of carbon and 2 atoms of oxygen are present in a molecule of carbon dioxide. Hence the formula of carbon dioxide is CO2.

120567.

Name the elements present in this gas and wire their valency. [Given, atomic masses of C = 12 u, O = 16 u, NA = 6.022 × 1023 per mole]

Answer»

The elements present in carbon dioxide gas are carbon and oxygen. The valency of carbon is 1 and the valency of oxygen is 2-

120568.

Convert into moles. a) 12 g of O2 gas b) 20 g of water c) 22 g of carbon dioxide

Answer»

a) 12 g of O2 gas : 

Molecular weight of O2 is 32. 

∴ Number of moles of 32 g of O2 gas = 1

Number of moles of 12 g of O gas = \(\frac{12}{32}\)  x 1 = 0.375

b) 20 g of water : 

Molecular weight of water (H2O) is 18. 

Number of moles of 18 g of water = 1 

Number of moles of 20 g of water = \(\frac{20}{18}\) x 1 = 1.11

c) 22 g of carbon dioxide : 

Molecular weight of carbon dioxide (CO2 ) is 44. 

∴ Number of moles of 44 g of CO2 = 1

Number of moles of 22 g of CO2 = \(\frac{22}{44}\) x 1 = \(\frac{1}{2}\)= 0.5

120569.

What is meant by gram atomic mass? Give example.

Answer»

Gram atomic mass of an element is its relative atomic mass expressed in grams 

Ex : (1)The gram atomic mass of a hydrogen atom is 1. 008 g. 

(2) The gram atomic mass of a oxygen atom is 16 g.

120570.

What is molecular mass of a substance ?

Answer»

The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of a substance.

120571.

Define the term atomic or molecular mass in grams.

Answer»

One mole of any species is the quantity that is expressed in a number having a mass equal to its atomic or molecular mass in grams.

120572.

Who proposed atomic theory ?

Answer»

Dalton proposed atomic theory.

120573.

What is atomic mass unit (AMU)?

Answer»

Atomic mass unit (AMU): One atomic mass unit is defined as mass exactly one-twelfth the atomic mass of carbon – 12 isotopes.

120574.

Define atomic mass.

Answer»

Atomic mass: The relative atomic mass of the atom of an element is defined as the average mass of the atom, as compared to 1/12th of the mass of one carbon – 12 atom.

120575.

Why is it not possible to see an atom with naked eyes?

Answer»

Atoms are very small, they are smaller than anything that we can imagine or compare with. Therefore it is not possible to see an atom with naked eyes.

120576.

Write the relative atomic masses of some elements.

Answer»
ElementSymbolRelative mass
AluminumAl26.98
CalciumCa40.08
CarbonC12.01
HydrogenH1.008
MagnesiumMg24.31
NitrogenN14.01
Oxygen015.99
SilverAg107.57
120577.

Calculate the relative molecular mass of carbon dioxide.

Answer»

Molecular formula of carbon dioxide =CO2 

Atomic mass of Carbon = 12 

Atomic mass of Oxygen = 16 

1 × atomic mass of carbon + 2 × atomic mass of oxygen. 

= (1 × 12) +(2 × 16) = 44 

Hence relative molecular mass of carbon dioxide = 44.

120578.

What is meant by relative molecular mass of a substance?

Answer»

The relative molecular mass of a substance is the ratio of the mass of a molecule of the substance to \(\frac{1}{12}\)the mass of an atom of Carbon 6C12 isotope.

120579.

Define the atomic mass unit.

Answer»

One atomic mass unit is a mass unit equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12.

120580.

Which postulate of Dalton’s theory can explain the law of constant proportions?

Answer»

Third postulate of Dalton’s theory i.e. “Atoms of a given element have identical mass and chemical properties. Atoms of different elements have different masses and chemical properties”, is the result of law of constant proportions.

120581.

Give a brief description of the following:(i) Relative atomic mass (ii) Atomic mass unit (iii) Ions (iv) Ionic compound (v) Atomicity

Answer»

(i) Relative atomic mass: The relative atomic mass of an element is deined as the average mass of the atom, as compared to 1/12th the mass of one-carbon atom. 

The relative atomic masses of all the elements have been found with respect to an atom of carbon-12. 

(ii) Atomic mass unit: One atomic mass unit is a mass unit equal to the exactly 1/12th the mass of one atom of carbon-12.

Carbon-112 was chosen as the standard reference for measuring atomic masses.

(iii) Ions: Compounds composed of metals and non-metals has charged species. The charge species are known as ions. The negatively charged ions are called anions and positively charged ions are called cations. 

(iv) Ionic compounds: Compounds formed by the exchange of ions (cations or anions) are called ionic compounds. 

In ionic compounds, the charge on each ion is used to determine the chemical formula of the compound.

(v) Atomicity: 

Atomicity: The number of atoms constituting a molecule is known as its atomicity. For example:

In P4 , there are four atoms of phosphorus constituting a molecule. Hence, the atomicity is 4 (tetra-atomic).

120582.

Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Answer»

Atoms combine in the ratio of small whole numbers to form compounds. This postulate explains the law of definite proportions.

120583.

Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Answer»

The relative number and kinds of atoms are constant in a given compound. This postulate is the result of the law of conservation of mass.

120584.

Write postulates of Dalton’s atomic theory.

Answer»

Dalton’s Atomic Theory 

Based upon laws of chemical combination, Dalton’s Atomic Theory provided an explanation for the Law of Conservation of Mass and Law of Constant Composition. 

Postulates of Dalton’s atomic theory are as follows : 

• All matter is made up of very tiny particles called ‘Atoms’. 

• Atom are indivisible particles, which can’t be created or destroyed in a chemical reaction. (Proves ‘Law of Conservation of Mass’) 

• Atoms of an element have identical mass and chemical properties. 

• Atoms of different elements have different mass and chemical properties. 

• Atom combine in the ratio of small whole numbers to form compounds. (Proves ‘Law of Constant Proportion’) 

• The relative number and kinds of atoms are constant in a given compound.

120585.

State the two important laws of chemical combination. How Dalton’s atomic theory explains the two Laws?

Answer»

The two important laws of chemical combination are:

i) Law of constant proportion: 

The law states that “in a chemical substance, the elements are always present in definite proportions by mass. 

For example: In ammonia (NH3), nitrogen and hydrogen are always present in the ratio of 14:3 by mass, whatever the method

or source from which it is obtained. 

In water, hydrogen and oxygen are always present in the ratio of 1:8 by mass, whatever the method or source from which water is obtained.

ii) Law of conservation of mass: 

According to the law of conservation of mass states, atoms are neither created nor destroyed in a chemical reaction. This means that the total mass of the products formed in a chemical reaction must be equal to the mass of reactants consumed.

For example: 

CaCO→ CaO + CO2

100g          56g     44g

Sum of mass of reactants = Sum of masses of prodcuts

120586.

a) State the various postulates of Dalton’s atomic theory of matter. (b) Which postulate of Dalton’s atomic theory can explain the law of conservation of mass? (c) Which postulate of Dalton’s atomic theory can explain the law of constant proportions?

Answer»

(a) Postulates of Dalton’s atomic theory: 

1) All the matter is made up of very small particles called ‘atoms’. 

2) Atoms cannot be divided. 

3) Atoms can neither be created nor be destroyed. 

4) Atoms are of various kinds. There are as many kinds of atoms as are elements. 

5) All the atoms of a given element are identical in every respect, having the same mass, size and chemical properties. 

6) Atoms of different elements differ in mass, size and chemical properties. 

7) The ‘number’ and ‘kind’ of atoms in a given compound is fixed. 

8) During chemical combination, atoms of different elements combine in small whole numbers to form compounds. 

9) Atoms of the same elements can combine in more than one ratio to form more than one compound. 

(b) The postulate “The elements consists of atoms and that atoms can neither be created nor destroyed” can be used to explain the law of conservation of mass. 

(c) The postulate “The elements consist of atoms having fixed mass, and that the number and kind of atoms of each element in a given compound is fixed” can be used to explain the law of constant proportions.

120587.

Which of the following represents 14 u?A. Mass of 1 hydrogen atomB. Mass of C-12 atomC. Mass of 0-16 atomD. 1/12th of mass of C-12 atom

Answer»

1/12th of mass of C-12 atom represents 14u. One atomic mass unit is a mass unit equal to one atom of carbon-12.

120588.

How Scientists compare the masses of different atoms of elements ?

Answer»

Scientists use the relative atomic mass scale to compare the masses of different atoms of elements.

120589.

Collect information about various organic acids different occurring naturally and prepare a table.

Answer»
1. Acetic acidVinegar (obtained from fruits after fermentation).
2. Citric acidCitrus fruits like orange and lemons.
3. Butyric acidButter gone bad or rancid
4. Lactic acidCurd
5. Malic acidApples
6. Oleic acidOlive oil
7. Tartaric acidFruits such as grapes, apples and tamarind
8. Stearic acidFrom fats
9. Succinic acidFrom vegetables like lettuce and unripe fruits
10. Uric acidFrom urine
120590.

Which of the following postulates of Dalton’s atomic theory is challenged by the next generation scientists?A. All matter is made of very tiny particles called atoms.B. Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.C. Atoms of a given element are identical in mass and chemical properties.D. Atoms of a given element are identical in mass and chemical properties.

Answer»

The postulate of Dalton’s atomic theory that “Atoms of a given element are identical in mass and chemical properties” was challenged by the next generation scientists. They have to prove this theory correct at any cost.

120591.

(i) Which element is used as a standard for atomic mass scale? (ii) What is the value of mass of this reference atom?

Answer»

i. In 1961, for a universally accepted atomic mass unit, carbon-12 isotope was chosen as the standard reference for measuring atomic masses.

ii. One atomic mass unit is a mass unit equal to exactly 1/12th the mass of one atom of carbon-12.

120592.

While searching for various atomic mass units, scientists initially took 1/16 of the mass of a naturally occurring oxygen atom as one unit. State two reasons for this.

Answer»

While searching for various atomic mass units, scientists initially took 1/16 of the mass of a naturally occurring oxygen atom as one unit. The two reasons are: 

i. According to them, oxygen reacted with a large number of elements and formed compounds. 

ii. The given atomic mass unit gave masses of most of the elements as whole numbers.

120593.

The opening and closing of spiracles are regulated by ………

Answer»

sphincter spiracular muscles

120594.

Which of the following lenses would you prefer to use while reading small letters found in a dictionary?(a) A convex lens of focal length 50 cm.(b) A concave lens of focal length 50 cm.(c) A convex lens of focal length 5 cm.(d) A concave lens of focal length 5 cm.

Answer»

(c) A convex lens of focal length 5cm.

120595.

A student takes 2ml acetic acid, in a dry test-tube and add a pinch of sodium hydrogen carbonate to it. Write down the observation made by the students ?

Answer»

A colourless and odourless gas is evolved. It turns lime water milky and this gas extinguishes the burning splinter

120596.

UHF परिसर की आवृत्तियों का प्रसारण प्रायः किसके द्वारा होता है? (a) भू-तरंगें (b) व्योम तरंगें (c) पृष्ठीय तरंगें (d) आकाश तरंगें

Answer»

(d) आकाश तरंगें। UHF परिसर में प्रसारण आकाश तरंगों द्वारा ही होता है।

120597.

व्योम तरंगों के उपयोग द्वारा क्षितिज के पार संचार के लिए निम्नलिखित आवृत्तियों में से कौन-सी आवृत्ति उपयुक्त रहेगी? (a) 10 kHz (b) 10 MHz (c) 1 GHz (d) 1000 GHz

Answer»

(b) 10 MHz 

3 MHz से 30 MHz आवृत्ति तक की तरंगें व्योम तरंगों की श्रेणी में आती हैं। इससे उच्च आवृत्ति की तरंगें (जैसे-1GHz, 1000 GHz) आयन मण्डल को भेदकर पार निकल जाती हैं जबकि 10 kHz आवृत्ति की तरंगें ऐन्टीना की ऊँचाई अधिक होने के कारण उपयोगी नहीं हैं।

120598.

What is a beam of light?

Answer»

A group of light rays originating from a source and travelling in some definite direction is known as a beam of light.

120599.

What is light?

Answer»

Light is a form of energy which gives us the sensation of sight or vision.

120600.

A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Answer»

Given P=+ 1.5D
As P=1/f (in metre) ,
Focal length, f =1/P=1/1.5=0.667m=66.7cm

As power of prescribed lens is positive, the lens is converging (or convex) lens.