Chemistry Test

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Chemistry Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

A solution containing $1.8 g$ of a compound (empirical formula ${C H}_{2} O$ ) in $40 g$ of water is observed to freeze at $- {0.465}^{\circ} C$ . The molecular formula of the compound is $(K_{f}$ of water $= 1.86 k g$ ${k m o l}^{- 1})$

A
$C_{2} H_{4} O_{2}$

B
$C_{3} H_{6}$

C
$C_{4} H_{8} O_{4}$

D
$C_{6} H_{12} O_{6}$

Solution

Let compound be B.

Water be A.

$Δ T_{f} = T_{f}^{0} - T_{F S}$

Here, $T_{f}^{0} = 0$

Therefore,

$Δ T_{f} = 0^{\circ} - (- 0.465)$

$Δ T_{f} = {0.465}^{\circ} C$

Given:

$K_{f} = 1.86 k g K m o l^{- 1}$

Weight of compound, $w_{B} = 1.8 g m$

Let Molecular weight of compound, $M_{B} = x$

Weight of water, $w_{A} = 40 g$

We know that:

$Δ T_{f} = K_{f} \cdot \frac{w_{B}}{M_{B}} \times \frac{1000}{w_{A}}$

$0.465 = \frac{1.86 \times 1.8 \times 1000}{x \times 40}$

$x = 180 g / m o l$

So, empirical mass of $(C H_{2} O)$ $= (12 \times 1) + (2 \times 1) + (16 \times 1)$

$= 30$

$n = \frac{Molecular Mass}{Empirical formula mass}$

$= \frac{180}{30}$

$n = 6$

So, molecular formula $= ($ Emperical formula $)_{n}$

$= {({C H}_{2} O)}_{6}$

$= C_{6} H_{12} O_{6}$
Question 2
1 / -0

Determine the empirical formula of an oxide of iron which has $69.9 %$ iron and $30.1 %$ dioxygen by mass.

A
${F e}_{2} O_{3}$

B
${F e}_{3} O_{2}$

C
${F e}_{3} O_{4}$

D
None of these

Solution

We know that:

The molar mass of oxygen $= 16$

The molar mass of iron $= 55.85$

Given:

Percentage of iron by mass $= 69.9 %$

Percentage of oxygen by mass $= 30.1 %$

Now,

Relative mass of iron in iron oxide $= \frac{Percentage of iron by mass}{At. mass of iron} = \frac{69.9}{55.85} = 1.25$

Relative mass of oxygen in iron oxide $= \frac{Percentage of oxygen by mass}{At. mass of oxygen} = \frac{30.1}{16} = 1.88$

Now,

Simplest molar ratio of iron to oxygen $= 1.25 : 1.88 = 2 : 3$

Therefore the empirical formula of iron oxide is ${F e}_{2} O_{3}$ .
Question 3
1 / -0

Hydrogen bond is strongest in:

A
F-H-O

B
F-H-N

C
F-H-F

D
All are equally strong

Solution

We know that, the strength of the hydrogen bond depends on the electronegativity of the atoms. In F-H-F there are two fluorine atoms which is most electronegative element. So, F-H-F has the strongest H-bond.
Question 4
1 / -0

Alkali metals give a _________________when dissolved in liquid ammonia.

A
deep blue solution

B
colourless

C
red colour

D
None of the Above

Solution

Alkali metals give a deep blue solution when dissolved in liquid ammonia.

When an alkali metal is dissolved in liquid ammonia, it results in the formation of a deep blue coloured solution. The ammoniated electrons absorb energy corresponding to a red region of visible light. Therefore, the transmitted light is blue in colour.
Question 5
1 / -0

The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is:

A
${Fe}_{4} {[Fe (CN)_{6}]}_{3}$

B
$N a_{3} [Fe (CN)_{6}]$

C
$Fe (CN)_{3}$

D
${Na}_{4} [Fe (CN)_{5} NOS]$

Solution

Lassaigen's test is a general method for determining the presence of halogens, phosphorus, and sulphur in an organic compound. The organic compound is fused with sodium metal in this process. The ionic compounds formed during fusion are collected in an aqueous solution and identified using basic chemical tests.

The carbon and nitrogen present in the organic compound on fusion with sodium metal give sodium cyanide $N a C N$ soluble in water.

$N a + C + N \to N a C N$

This is converted into sodium ferrocyanide by the addition of sufficient quantities of ferrous sulphate.

$6 NaCN + {FeSO}_{4} \to {Na}_{4} [Fe (CN)_{6}] + {Na}_{2} {SO}_{4}$

The ferric ions generated during the process react with ferrocyanide to form a Prussian blue ferric ferrocyanide precipitate.

${Na}_{4} [Fe (CN)_{6}] + {Fe}^{3 +} \to {Fe}_{4} {[Fe (CN)_{6}]}_{3}$

Ferric ferrocyanide turns Prussian blue in this reaction, hence it shows a positive test for nitrogen with Lassaigne's solution.
Question 6
1 / -0

Which of the following is not a homologous series?

A
${CH}_{4}$

B
$C_{2} H_{6}$

C
$C_{4} H_{9}$

D
$C_{3} H_{8}$

Solution

$C_{4} H_{9}$ does not belongs to homologous series.

Homologous series is a series of compounds with similar chemical properties and some functional groups differing from the successive member by ${CH}_{2}$ .

Carbon chains of varying length have been observed in organic compounds having the same general formula.

Alkanes with general formula $C_{n} H_{2 n + 2}$ alkenes with general formula $C_{n} H_{2 n}$ and alkynes with general formula $C_{n} H_{2 n - 2}$ form the most basic homologous series in organic chemistry.
Question 7
1 / -0

The result of the operation $2.5 \times 1.25$ should be which of the following on the basis of significant figures?

A
$3.125$

B
$3.13$

C
$3.1$

D
$31.25$

Solution

Here,

$2.5 \times 1.25 = 3.125$

The rule says: Answer to a multiplication or division should be rounded off to a same number of significant figures as possessed by the least precise term in the calculation.

Since $2.5$ has least numbers of significant figure i.e., two, thus, the result should have 2 significant figure i.e., $3.1$ .
Question 8
1 / -0

Paramagnetism is common in:

A
s-block elements

B
p-block elements

C
f-block elements

D
None of these

Solution

Paramagnetism is common ins-block elements.

s-block element has a single electron in its valence shell. The single unpaired electron is responsible for the paramagnetism property.Paramagnetism is a form of magnetism where some materials are weekly attracted by an externally applied magnetic field.
Question 9
1 / -0

$1 kg$ of 0.75 molal aqueous solution of sucrose can be cooled up to $- 4^{\circ} C$ before freezing. The amount of ice (in g) that will be separated out is__________(Nearest integer)

[Given, $K_{f} (H_{2} O) = 1.86 Kkg {mol}^{- 1}$ ]

A
452

B
518

C
315

D
625

Solution

$\begin{aligned} Let mass of water (initially present) = xg \\ Mass of sucrose = (1000 - x) g \\ Moles of sucrose = \frac{1000 - x}{342} \\ Molality = \frac{moles of sucrose}{mass of water(initially)} \\ 0.75 = \frac{(\frac{1000 - x}{342})}{(\frac{x}{1000})} \\ \frac{x}{1000} = \frac{1000 - x}{342 \times 0.75} \\ 256.5 x = 10^{6} - 1000 x \\ \Rightarrow x = 795.86 g \\ Moles of sucrose = 0.5969 \\ New mass of H_{2} O = a kg \\ Depression in freezing point Δ T_{f} = K_{f} \times m \\ 4 = \frac{0.5969}{a} \times 1.86 \\ \Rightarrow a = 0.2775 kg \\ Ice separated = 795.86 - 277.5 \\ = 518.3 g \approx 518 g . \end{aligned}$
Question 10
1 / -0

In liquids London forces are found in:

A
Polar molecules

B
Non-polar molecules

C
Hydrogen bonded molecules

D
None of the above

Solution

Intermolecular forces are the attractive and repulsive forces that arise between the molecules of a substance.

Intermolecular forces are mainly responsible for the physical characteristics of the substance.

The stronger the noncovalent interactions between molecules, the more energy that is required, in the form of heat, to break them apart.

The boiling point and melting point of a substance is proportional to the strength of its non-covalent intermolecular forces i.e. the stronger the intermolecular forces, the higher the boiling point and melting point.

The strongest intermolecular force is the hydrogen bond.

The weakest intermolecular forces are London dispersion forces.

London Forces:

The instantaneous dipole–induced dipole attractions are called London dispersion forces.

The name is given after Fritz London (1900–1954), a German physicist who developed this model to explain the intermolecular attractions that exist between non-polar molecules.

London’s dispersion forces occur between all molecules.

These very weak attractions occur because of the random motions of electrons on atoms within molecules.

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Chemistry Test - 34

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Chemistry Test - 34

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Question 1

C2H4O2

C3H6

C4H8O4

C6H12O6

Solution

Question 2

Fe2O3

Fe3O2

Fe3O4

None of these

Solution

Question 3

F-H-O

F-H-N

F-H-F

All are equally strong

Solution

Question 4

deep blue solution

colourless

red colour

None of the Above

Solution

Question 5

Fe4[Fe(CN)6]3

Na3[Fe(CN)6]

Fe(CN)3

Na4[Fe(CN)5NOS]

Solution

Question 6

CH4

C2H6

C4H9

C3H8

Solution

Question 7

3.125

3.13

3.1

31.25

Solution

Question 8

s-block elements

p-block elements

f-block elements

None of these

Solution

Question 9

452

518

315

625

Solution

Question 10

Polar molecules

Non-polar molecules

Hydrogen bonded molecules

None of the above

Solution

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$C_{2} H_{4} O_{2}$

$C_{3} H_{6}$

$C_{4} H_{8} O_{4}$

$C_{6} H_{12} O_{6}$

${F e}_{2} O_{3}$

${F e}_{3} O_{2}$

${F e}_{3} O_{4}$

${Fe}_{4} {[Fe (CN)_{6}]}_{3}$

$N a_{3} [Fe (CN)_{6}]$

$Fe (CN)_{3}$

${Na}_{4} [Fe (CN)_{5} NOS]$

${CH}_{4}$

$C_{2} H_{6}$

$C_{4} H_{9}$

$C_{3} H_{8}$

$3.125$

$3.13$

$3.1$

$31.25$