Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

According to the law of conservation of mass, a balanced chemical equation

A
the same number of electrons of each element on both sides of the equation

B
the same number of atoms of each element on both sides of the equation

C
the same ratio of atoms of each element on both sides of the equation

D
the same number of protons of each element on both sides of the equation

Solution

According to law of conservation of mass

total mass of reactants

= total mass of product

∴∴ the number . of atoms of each element in reactants

= number . of atoms of that element in products
Question 2
1 / -0

Molarity is defined as,

A
the number of moles of the solute in 1 litre of the solution

B
the number of moles of the solute in 1 m3 of the solution

C
the number of grams of the solute in 1 litre of the solution

D
the number of moles of the solvent in 1 litre of the solution

Solution

Molarity is the most widely used method of expressing the strength or concentration of a solution .

It is denoted by ' M ' .

Molarity is defined as number of . of moles of solute present per litre of solution.

Mathematically ,

Molarity ( M )

= Number of moles of solute / Volume of solution in Litres
Question 3
1 / -0

The calculation of masses or, (sometimes volumes also) of the reactants and the products involved in a chemical reaction is called

A
stoichiometry

B
molality

C
normality

D
molarity

Solution

Stoichiometry is a method to express quantitative aspects of  a chemical reaction .

Usually , the masses of reactants as well as those of products in a chemical reaction are calculated using corresponding  balanced chemical equation .

It is convenient and hence  desirable to calculate volumes of gaseous reactants and products .
Question 4
1 / -0

Molecular mass of glucose (C₆H₁₂O₆) is

A
180. 162 u

B
192.162 u

C
198.162 u

D
206.162 u.

Solution

Molecular mass of Glucose is calculated using the relation :

Molecular mass

= ∑ ( 6*atomic mass of C , 12* atomic mass of H , 6* atomic mass of O ) u

∴∴ substituting the respective atomic masses we get ,

Molecular mass of glucose ( C₆ H₁₂O₆ )

= [ 6(12.0107)+12(1.008)+6(15.9994) ]

= 180.162 u.
Question 5
1 / -0

Choose the most appropriate answer amongst the options given below for the statement -

" A solution of a desired concentration is prepared by diluting --------------- ."

A
solution of known lower concentration

B
stock solution.

C
from a serially diluted solution

D
solution of known higher concentration

Solution

Stock solution can best be described as concentrated solution of known accurate concentration that will be used for future laboratory use.

Since large amounts of solutes are used for preparing stock solution a more accurate concentration of it can be achieved quite easily, and as such the chances are slim to get erroneous results of the related experiments..

In addition , stock solutions are generally more stable as compared to working solution since they usually do not suppport bacterial growth.

A multiple working solution can be prepaed by dilution of stock solution using easy calculation and process.
Question 6
1 / -0

The number of moles of solute present in 1 kg of solvent is called--------------.

A
molarity (M)

B
ppm

C
molality (m)

D
Normality( N )

Solution

Molality is defined as no. of moles of solute present per kg of solvent.It is denoted as " m " . Mathematically ,

Molality (m)

= [ number of moles of the solute ] / kg of solvent

Since , this mode of expressing the strength of a solution involves ( weight / weight ) relationship of solute and solvent , the molality of the solution is not affected by variation in temperatures of the solution.
Question 7
1 / -0

The density of 3 M solution of NaCl is1.25 g mL-¹ . The molality of the solution is ,

A
2.79 m

B
1.79 m

C
2.09 m

D
0.79 m

Solution

Calculations :

Since , Molarity represents number of moles of soute per litre of solution .

∴∴ moles of solute

= 3 mols.

As , density of the solution is given

= 1.25 g / mL

∴∴

Mass of solution

=1250g

Mass of solvent =

[1250 - 3 (58.5)] g

= ( 1250 - 175.5 ) g

=1074 .5 g

=1.0745 Kg

Molality

= Number of moles of solute / Kg of solvent

∴∴Molality

= ( 3 mol / 1.0745 kg )

=2.79 m
Question 8
1 / -0

0.2M NaOH means

A
0.2 moles of NaOH in 0.50 lires of solution

B
0.2 moles of NaOH in 10 mL of solution

C
0.2 moles of NaOH per lire of solution

D
0.2 moles of NaOH in 100 mL of solution

Solution

Since , M denotes molarity of a solution & Molarity

= number of moles of a solute (ie given as NaOH ) / Volume of solution in Litres

∴. 0 . 2M solution means 0.2moles of solute(NaOH) present in 1L of solution.
Question 9
1 / -0

A solution is prepared by adding 2 g of a substance A to 18 g of water.Mass per cent of the solute is ---------------.

A
11 %

B
20.5 %

C
10 %

D
1 2.11%

Solution

Calculations :

Since , mass %

=[ { (mass of solute) x) / (mass of solution ) } x 100 ]

∴∴ Substituting the given values we get,

Mass per cent

= [ ( 2 g ) / { (2 g of A ) + 18 g of water } ]

= [ ( 2 /20 ) x 100 ]

= 10%

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Some Basic Concepts of Chemistry Test - 7

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Some Basic Concepts of Chemistry Test - 7

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Question 1

the same number of electrons of each element on both sides of the equation

the same number of atoms of each element on both sides of the equation

the same ratio of atoms of each element on both sides of the equation

the same number of protons of each element on both sides of the equation

Solution

Question 2

the number of moles of the solute in 1 litre of the solution

the number of moles of the solute in 1 m3 of the solution

the number of grams of the solute in 1 litre of the solution

the number of moles of the solvent in 1 litre of the solution

Solution

Question 3

stoichiometry

molality

normality

molarity

Solution

Question 4

180. 162 u

192.162 u

198.162 u

206.162 u.

Solution

Question 5

solution of known lower concentration

stock solution.

from a serially diluted solution

solution of known higher concentration

Solution

Question 6

molarity (M)

ppm

molality (m)

Normality( N )

Solution

Question 7

2.79 m

1.79 m

2.09 m

0.79 m

Solution

Question 8

0.2 moles of NaOH in 0.50 lires of solution

0.2 moles of NaOH in 10 mL of solution

0.2 moles of NaOH per lire of solution

0.2 moles of NaOH in 100 mL of solution

Solution

Question 9

11 %

20.5 %

10 %

1 2.11%

Solution

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