Some Basic Concepts of Chemistry — Page 7 | NEET Chemistry

Q61

Complete combustion of 1.80 g of an oxygen containing compound (CxHyOz) gave 2.64 g of CO2 and 1.08 g of H2O. The percentage of oxygen in the organic compound is :

A 51.63

B 50.33

C 63.53

D 53.33

Correct Answer

Option D

Solution

CxHyOz + O2 $\to$ xCO2 +

{y \over 2}{H_2}O

2.64 g of CO2 contains 0.72 g C.

1.08 g of H2O contains 0.12 g H.

$\therefore$ Mass of oxygen present = 1.80 – (0.72 +0.12) = 0.96 g % of O =

{{0.96} \over {1.80}} \times 100

= 53.33 %

Q62

Excess of NaOH (aq) was added to 100 mL of FeCl3 (aq) resulting into 2.14 g of Fe(OH)3 . The molarity of FeCl3 (aq) is : (Given molar mass of Fe = 56 g mol−1 and molar mass of Cl = 35.5 g mol−1)

A 0.2 M

B 03 M

C 0.6 M

D 1.8 M

Correct Answer

Option A

Solution

3 NaOH (aq.) + FeCl3(aq) $\to$ Fe(OH)3(s)+ 3 NaCl(aq). Moles of Fe(OH)3 =

{{2.14} \over {107}}

= 2 $\times$ 10 $-$ 2 1 mole of Fe(OH)3 is obtained from = 1 mole of FeCl3

\therefore\,\,\,

2 $\times$ 10 $-$ 2 moles of Fe(OH)3 will obtain from = 0.02 mole of FeCl3 Molarity of FeCl3 =

{{No.of\,moles} \over {Volume\,in\,L}}

=

{{2 \times {{10}^{ - 2}}} \over {0.1}}

= 0.2 M

Q63

On combustion 0.210 g of an organic compound containing C, H and O gave 0.127 g H2O and 0.307 g CO2. The percentages of hydrogen and oxygen in the given organic compound respectively are:

A 7.55, 43.85

B 6.72, 53.41

C 6.72, 39.87

D 53.41, 39.6

Correct Answer

Option B

Solution

In the combustion of organic compound, all "C" in $\mathrm{CO}_2$ and all " H " in $\mathrm{H}_2 \mathrm{O}$ comes from organic compound Weight of " C " in $\mathrm{CO}_2=\dfrac{12}{44} \times 0.307$

=0.0837 \mathrm{~gm}

Weight of " H " in $\mathrm{H}_2 \mathrm{O}=\dfrac{2}{18} \times 0.127=0.0141 \mathrm{~g}$

\begin{aligned} &\begin{gathered} \% ' H^{\prime} \text{ in compound }=\frac{0.0141}{0.21} \times 100=6.719 \% \\ =6.72 \% \end{gathered}\\ &\text{ Weight of "O" in compound }\\ &\begin{aligned} & =0.210-(0.0837+0.0141) \\ & =0.1122 \end{aligned}\\ &\begin{aligned} & \% \text{ of "O" in compound }=\frac{0.1122}{0.21} \times 100 \\ & =53.41 \% \end{aligned} \end{aligned}

Q64

The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCl will be:

A 1.00 M

B 1. 75 M

C 0.975 M

D 0.875 M

Correct Answer

Option D

Solution

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} = 0.5

,

{V_1} = 750

,

{M_2} = 2

,

{V_2} = 250

$\therefore$ Molarity of mixture =

{{0.5 \times 750 + 2 \times 250} \over {750 + 250}}

=

{{375 + 500} \over {1000}}

=

{{875} \over {1000}}

= 0.875

Q65

The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid istreated with excess H2S in the presence of conc. HCl ( assuming 100% conversion) is :

A 0.50 mol

B 0.25 mol

C 0.125 mol

D 0.333 mol

Correct Answer

Option C

Solution

2H3 As O4 + 5H2S

\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{HCl}^{conc}}

As2S5 + 8H2O 35.5 g of H3AsO4 =

{{35.5} \over {142}}

= 0.25 moles Let, As2S5 produced = n moles.

\therefore\,\,\,

{{0.25} \over 2} = {n \over 1}

$\Rightarrow$

\,\,\,

n = 0.125 mol.

Q66

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 }

Q67

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

Q68

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.

Q69

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.

Q70

The number of molecules and moles in 2.8375 litres of O

_2

at STP are respectively

A 1.505

\times

10

^{23}

and 0.250 mol

B 7.527

\times

10

^{22}

and 0.250 mol

C 7.527

\times

10

^{23}

and 0.125 mol

D 7.527

\times

10

^{22}

and 0.125 mol

Correct Answer

Option D

Solution

At STP, one mole of any gas occupies 22.4 liters.

Therefore, 2.8375 liters of oxygen gas at STP is equal to 0.125 moles.

The number of molecules of oxygen gas in 0.125 moles is calculated by multiplying the number of moles by Avogadro's number, which is

6.022 × 10^{23}

molecules/mol. This gives us a total of

7.527 × 10^{22}

molecules of oxygen gas in 2.8375 liters at STP. Here is the calculation:

\begin{align} &\text{Number of moles of }O_2 = \frac{2.8375 \text{ L}}{22.4 \text{ L/mol}} = 0.125 \text{ mol} \\\\ &\text{Number of molecules of }O_2 = 0.125 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 7.527 \times 10^{22} \text{ molecules} \end{align}