Some Basic Concepts of Chemistry — Page 7 | JEE Chemistry

Q61

The density of '

x

'

\mathrm{M}

solution ('

x

' molar) of

\mathrm{NaOH}

is

1.12 \mathrm{~g} \mathrm{~mL}^{-1}

, while in molality, the concentration of the solution is

3 \mathrm{~m}

( 3 molal). Then

x

is (Given : Molar mass of

\mathrm{NaOH}

is

40 \mathrm{~g} / \mathrm{mol}

)

A 3.8

B 3.0

C 3.5

D 2.8

Correct Answer

Option B

Solution

To find the molarity,

x

, of the

\mathrm{NaOH}

solution, we need to relate the given density and molality.

Molarity is defined as the number of moles of solute per liter of solution, while molality is defined as the number of moles of solute per kilogram of solvent.

We are given the following: Molality (m) =

3\ \mathrm{mol/kg}

(3 molal) Density of solution ( $\rho$ ) =

1.12\ \mathrm{g/mL}

=

1120\ \mathrm{g/L}

Molar mass of

\mathrm{NaOH}

(

M_\mathrm{NaOH}

) =

40\ \mathrm{g/mol}

First, let's find the number of moles of

\mathrm{NaOH}

in 1 kg of water, which is the definition of molality. Since the molality is 3 molal, we have:

3\ \mathrm{mol}

of

\mathrm{NaOH}

in 1 kg of water (or 1000 g of water).

Now, to find the molarity, we need the volume of the solution in which these moles of

\mathrm{NaOH}

are dissolved. The mass of 1 L (1000 mL) of solution, given the density, is:

1120\ \mathrm{g}

The mass of

\mathrm{NaOH}

in 3 moles is:

3 \mathrm{mol} \times 40\ \mathrm{g/mol} = 120\ \mathrm{g}

Thus, the mass of water in this 1 L solution can be found by subtracting the mass of

\mathrm{NaOH}

from the total mass of the solution:

1120\ \mathrm{g} - 120\ \mathrm{g} = 1000\ \mathrm{g}

So, we have 3 moles of

\mathrm{NaOH}

in 1 L of solution.

Therefore, the molarity (x) of the solution is directly 3 M (since molarity is moles of solute per liter of solution).

Hence,

x = 3

. Therefore, the correct answer is Option B: 3.0.

Q62

The quantity which changes with temperature is :

A Molality

B Molarity

C Mole fraction

D Mass percentage

Correct Answer

Option B

Solution

The quantity that changes with temperature from the given options is Molarity.

Molarity, which is denoted with a capital M, measures the number of moles of solute dissolved in one liter of solution.

It is expressed as:

Molarity (M) = \frac{moles \ of \ solute}{liters \ of \ solution}

Since the volume of liquids expands or contracts with temperature, the volume of the solution will change as the temperature changes, which in turn will affect the molarity.

For example, if the temperature of a solution increases, its volume may expand.

If the number of moles of solute remains constant, the increased volume will result in a decreased molarity because the solute is now dispersed in a greater volume of solution.

In contrast, other measures of concentration such as: Molality, which is measured as the number of moles of solute per kilogram of solvent (not affected by temperature since it's based on mass, not volume),

Molality (m) = \frac{moles \ of \ solute}{kilograms \ of \ solvent}

Mole fraction, which is the ratio of the number of moles of one component to the total number of moles of all components in the solution (also not affected by temperature since the ratio of moles doesn't change with temperature),

Mole \ fraction \ (X) = \frac{moles \ of \ component}{total \ moles \ in \ mixture}

and Mass percentage, which is the percentage by mass of a component in a mixture (again, not affected by temperature because it's based on mass),

Mass \ percentage \

(%)

= \frac{mass \ of \ component}{total \ mass \ of \ mixture} \times 100\%

do not change with temperature. Therefore, the correct answer to the question is: Option B : Molarity

Q63

How many moles of magnesium phosphate Mg3(PO4)2 will contain 0.25 mole of oxygen atoms?

A 1.25

\times

10-2

B 2.5

\times

10-2

C 0.02

D 3.125

\times

10-2

Correct Answer

Option D

Solution

In one Mg3(PO4)2 molecule no of oxygen atoms = 8 So if there is M mole of Mg3(PO4)2 molecules then no of moles of oxygen atoms is = M $\times$ 8 According to question, M $\times$ 8 = 0.25 $\Rightarrow$ M = 0.03125 = 3.125 $\times$ 10-2

Q64

An organic compound contains C, H and S. The minimum molecular weight of the compound containing 8% sulphur is : (atomic weight of S = 32 amu)

A 200 g mol

-

1

B 400 g mol

-

1

C 600 g mol−1

D 300 g mol−1

Correct Answer

Option B

Solution

We know that 8% sulphur means 8 g of sulphur present in 100 g of an organic compound. Thus, 32 g of sulphur present in

{{100} \over 8} \times 32g

= 400 g of organic compound. Therefore, minimum molecular weight of the compound is 400 g mol $-$ 1.

Q65

The elemental composition of a compound is

54.2 \% \mathrm{C}, 9.2 \% \mathrm{H}

and

36.6 \% \mathrm{O}

. If the molar mass of the compound is

132 \mathrm{~g} \mathrm{~mol}^{-1}

, the molecular formula of the compound is : [Given : The relative atomic mass of

\mathrm{C}: \mathrm{H}: \mathrm{O}=12: 1: 16

]

A

\mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_2

B

\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6

C

\mathrm{C}_4 \mathrm{H}_9 \mathrm{O}_3

D

\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_3

Correct Answer

Option D

Solution

\begin{array}{lllll} \mathrm{C} & : & \mathrm{H} & : & \mathrm{O} \\ \frac{54.2}{12} & : & 9.2 & : & \frac{36.6}{16} \\ 4.516 & : & 9.2 & : & 2.287 \\ \frac{4.516}{2.287} & : & \frac{9.2}{2.287} & : & \frac{2.287}{2.287} \\ 1.97 & : & 4.02 & : & 1 \end{array}

$\mathrm{C}_2 \mathrm{H}_4 \mathrm{O} \Rightarrow$ Empirical formula E.F. mass $=24+4+16=44$ and molar mass $=132$

\begin{aligned} \text{ Hence molecular formula } & =\left(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}\right)_3 \\ & =\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_3 \end{aligned}

Correct Option (3)

Q66

In a compound C, H and N atoms are present in 9 : 1 : 3.5 by weight . Molecular weight of the compound is 108 g mol-1 . Molecular formula of compound is

A C2H6N2

B C3H4N

C C6H8N2

D C9H12N3

Correct Answer

Option C

Solution

Element Weight ratio No of atoms of element Simplest Ratio C 9 9/12 = 3/4 3 H 1 1/1 = 1 4 N 3.5 3.5/14 = 1/4 1 Empherical Formula of the compound (EF) will be C3H4N Now to find molecular formula at first find number of moles of given mass No. of moles (n) = Given mass Molar mass $\Rightarrow$

{{108} \over {54}}

$\Rightarrow$ 2 $\therefore$ No. of Moles is 2 Molecular formula (MF) = n (no. of moles) $\times$ Empherical Formula (EF) $\Rightarrow$ 2 $\times$ C3H4N $\Rightarrow$ C6H8N2 $\therefore$ Option (C) is correct

Q67

Concentrated nitric acid is labelled as

75 \%

by mass. The volume in mL of the solution which contains 30 g of nitric acid is ______________. Given : Density of nitric acid solution is

1.25 \mathrm{~g} / \mathrm{mL}

.

A 32

B 40

C 55

D 45

Correct Answer

Option A

Solution

$\% \mathrm{w} / \mathrm{w}$ of $\mathrm{HNO}_3=75 \%$ means 100 gm of solution containing 75 g of $\mathrm{HNO}_3$ $\&\left(\dfrac{\mathrm{gm}}{\mathrm{m}_1}\right)_{\text{solution }}=1.25=\dfrac{100 \mathrm{gm}}{\mathrm{V}}$ $\mathrm{V}_{\mathrm{ml}}$ of 100 gm solution $=\dfrac{100}{1.25} \mathrm{ml}$ $\because 75 \mathrm{gm}$ of $\mathrm{HNO}_3$ present in $\dfrac{100}{1.25} \mathrm{ml}$ solution $\therefore 30 \mathrm{gm}$ of $\mathrm{HNO}_3$ present in

\frac{100}{1.25 \times 75} \times 30=32 \mathrm{ml} \text{ solution }

Q68

Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 ml of 1.2 M second solution. What is the molarity of the final mixture?

A 2.70 M

B 1.344 M

C 1.50 M

D 1.20 M

Correct Answer

Option B

Solution

Short Cut Method : Molarity, Normality, Average Atomic Mass, Average Molar Mass, Average Density, Average Vapour Pressure all of those values will be between the solutions which are going to be mixed.

Here molarity of first solution is 1.2 and molarity of second solution is 1.5.

So the molarity of the mixture will always be more than 1.2 and less than 1.5.

Remember the molarity of mixture can't be either 1.2 or 1.5.

So from the option you an see only 1.344 M can be the right answer.

Normal method : The formula for molarity of mixture of two substance is =

{{{M_1}{V_1} + {M_2}{V_2}} \over {{V_1} + {V_2}}}

Here

{M_1} = 1.5

,

{V_1} = 480

,

{M_2} = 1.2

,

{V_2} = 520

$\therefore$ Molarity of mixture =

{{1.5 \times 480 + 1.2 \times 520} \over {480 + 520}}

=

{{720 + 624} \over {1000}}

=

{{1344} \over {1000}}

= 1.344

Q69

Amongst the following statements, that which was not proposed by Dalton was :

A Matter consists of indivisible atoms all the atoms of a given element have.

B Chemical reactions involve reorganization of atoms. These are neither created not destroyed in a chemical reaction.

C When gases combine or reproduced in a chemical reactionn they do so in a simple ratio by volume provided all gases are the same T & P.

D Identical properties including identical mass. Atoms of differemt element differ in mass.

Correct Answer

Option C

Solution

Option(3) is according to Avogadro's law of volume combination.

Q70

A metal chloride contains

55.0 \%

of chlorine by weight .

100 \mathrm{~mL}

vapours of the metal chloride at STP weigh

0.57 \mathrm{~g}

. The molecular formula of the metal chloride is (Given: Atomic mass of chlorine is

35.5 \mathrm{u}

)

A

\mathrm{MCl}_{2}

B

\mathrm{MCl}_{4}

C

\mathrm{MCl}_{3}

D

\mathrm{MCl}

Correct Answer

Option A

Solution

The weight percent of chlorine in the compound is given as 55.0%.

This implies that the weight percent of the metal is 45.0%.

Now, we need to find the molar mass of the compound.

At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 L.

We know that 100 mL (or 0.1 L) of the metal chloride vapor weighs 0.57 g.

Therefore, the molar mass of the compound (MClx) is:

M_{MCl_x} = \frac{0.57 \, \text{g}}{0.1 \, \text{L}} \times 22.4 \, \text{L/mol} = 127.68 \, \text{g/mol}

Given the weight percent of the metal and chlorine, we can find the molar mass of the metal (MM) using the equation:

MM = M_{MCl_x} \times \frac{45.0}{55.0} = 104.04 \, \text{g/mol}

Subtracting the atomic mass of chlorine from the molar mass of the metal gives the molar mass of the metal:

MM - x \cdot 35.5 = 104.04 \, \text{g/mol}

Where 'x' is the number of chlorine atoms in the compound. Solving this equation for 'x' gives:

x = \frac{104.04 - MM}{35.5}

We find that 'x' is approximately 2.

Therefore, the molecular formula of the compound is MCl2.

So, the correct answer is MCl2.