Some Basic Concepts of Chemistry — Page 5 | JEE Chemistry

Q41

Density of 3 M NaCl solution is

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

. The molality of the solution is :

A 2.79 m

B 2 m

C 1.79 m

D 3 m

Correct Answer

Option A

Solution

\text{Molarity of NaCl} = 3 \text{ M} \quad (\text{3 moles in 1 L of solution})

Given the density of the solution is

1.25 \, \text{g/mL},

the total mass of 1 liter (1000 mL) of the solution is:

1.25 \, \text{g/mL} \times 1000 \, \text{mL} = 1250 \, \text{g}

To find the molality, we need the mass of the solvent (water).

First, calculate the mass of NaCl in 1 L of solution.

The molar mass of NaCl is approximately:

23 \, \text{g/mol (Na)} + 35.45 \, \text{g/mol (Cl)} \approx 58.45 \, \text{g/mol}

Thus, the mass of NaCl is:

3 \, \text{mol} \times 58.45 \, \text{g/mol} \approx 175.35 \, \text{g}

Now, the mass of the solvent (water) is:

1250 \, \text{g (solution)} - 175.35 \, \text{g (NaCl)} \approx 1074.65 \, \text{g}

Convert the mass of the solvent to kilograms:

1074.65 \, \text{g} \div 1000 \approx 1.07465 \, \text{kg}

Molality ( $m$ ) is defined as the number of moles of solute per kilogram of solvent:

m = \frac{3 \, \text{mol}}{1.07465 \, \text{kg}} \approx 2.79 \, \text{mol/kg}

Thus, the molality of the solution is approximately

2.79 \, \text{m}.

Q42

2.8 \times 10^{-3} \mathrm{~mol}

of

\mathrm{CO}_2

is left after removing

10^{21}

molecules from its '

x

' mg sample. The mass of

\mathrm{CO}_2

taken initially is Given:

\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}

A 98.3 mg

B 196.2 mg

C 150.4 mg

D 48.2 mg

Correct Answer

Option B

Solution

\begin{aligned} & (\text{ moles })_{\text{initial }}=\frac{x \times 10^{-3}}{44} \\ & (\text{ moles })_{\text{removal }}=\frac{10^{21}}{6.02 \times 10^{23}} \\ & (\text{ moles })_{\text{left }}=(\text{ moles })_{\text{initial }}-(\text{ moles })_{\text{removed }} \\ & 2.8 \times 10^{-3}=\frac{x \times 10^{-3}}{44}-\frac{10^{21}}{6.02 \times 10^{23}} \\ & \Rightarrow x=196.2 \mathrm{mg} \end{aligned}

Q43

Among

10^{-9} \mathrm{~g}

(each) of the following elements, which one will have the highest number of atoms? Element:

\mathrm{Pb}, \mathrm{Po}, \mathrm{Pr}

and Pt

A Pr

B Pt

C Po

D Pb

Correct Answer

Option A

Solution

To determine which element has the highest number of atoms among 10-9 grams of the elements Pb, Po, Pr, and Pt, we use the formula for calculating the number of atoms: $\text{Number of atoms} = \dfrac{\text{Mass}}{\text{Molar Mass (g/mol)}} \times N_A$ where $N_A$ is Avogadro's number.

From this formula, it's clear that for a given mass, the element with the smallest molar mass will have the greatest number of atoms, as it appears in the denominator.

Below are the molar masses of the elements: Molar mass of Pr (Praseodymium), $M_{\text{Pr}} = 141$ g/mol Molar mass of Pt (Platinum), $M_{\text{Pt}} = 195$ g/mol Molar mass of Pb (Lead), $M_{\text{Pb}} = 207$ g/mol Molar mass of Po (Polonium), $M_{\text{Po}} = 209$ g/mol Since Praseodymium (Pr) has the smallest molar mass ( $M_{\text{Pr}} = 141$ g/mol), 10-9 grams of Pr will contain the highest number of atoms compared to the other elements listed.

Q44

Mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dil. HCl is Given: Molar mass of Mg is

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

.

A 0.24 mg

B 235.7 g

C 2.444 g

D 236 mg

Correct Answer

Option D

Solution

We are tasked with calculating the mass of magnesium required to produce 220 mL of hydrogen gas at STP with excess dilute HCl.

Step 1: Write the balanced reaction $\text{Mg} + 2 \text{HCl} \;\;\rightarrow\;\; \text{MgCl}_2 + \text{H}_2$ 1 mole Mg gives 1 mole H₂.

Step 2: Molar volume at STP At STP, 1 mole of an ideal gas occupies 22.4 L = 22,400 mL.

Step 3: Moles of H₂ produced $n(\text{H}_2) = \dfrac{\text{Volume of H₂}}{\text{Molar Volume}} = \dfrac{220}{22400} = 0.00982 \,\text{mol}$ Step 4: Moles of Mg required From stoichiometry, 1 mole Mg produces 1 mole H₂, so: $n(\text{Mg}) = n(\text{H}_2) = 0.00982 \,\text{mol}$ Step 5: Mass of Mg required $m(\text{Mg}) = n \times M(\text{Mg}) = 0.00982 \times 24 = 0.236 \,\text{g}$ $= 236 \,\text{mg}$ ✅ Final Answer: Option D: 236 mg

Q45

At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O2 for complete combustion, and 40 mL of CO2 is formed. The formula of the hydrocarbon is :

A C4H7Cl

B C4H6

C C4H8

D C4H10

Correct Answer

Option B

Solution

CxHy +

\left( {x + {y \over 4}} \right)

O2 $\to$ xCO2 +

{y \over 2}

H2O According to the above equation, 1 mL of hydrocarbon = x mL of CO2 is produced $\therefore$ 10 mL of hydrocarbon = 10x mL of CO2 is produced.

According to question, From 10 mL of hydrocarbon 40 mL of CO2 is produced. $\therefore$ 10x = 40 $\Rightarrow$ x = 4 Also from the above equation, 1 mL of CxHy react with

\left( {x + {y \over 4}} \right)

mL of O2. $\therefore$ 10 mL of CxHy react with 10

\left( {x + {y \over 4}} \right)

mL of O2. According to question, 10

\left( {x + {y \over 4}} \right)

= 55 $\Rightarrow$

40 + {{10y} \over 4} = 55

$\Rightarrow$ y = 6 $\therefore$ Hydrocarbon is C4H6

Q46

1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar mass of M2CO3 in g mol–1 is:

A 84.3

B 118.6

C 11.86

D 1186

Correct Answer

Option A

Solution

M2CO3 + 2HCl

\overset{\,}\longrightarrow

2MCl + CO2 + H2O Here weight of M2CO3 = 1 gm Let molar mass of M2CO3 = M

\therefore\,\,\,

No of moles of M2CO3 =

{1 \over M}

For this reaction,

{{{1 \over M}} \over 1} = {{0.01186} \over 1}

\Rightarrow \,\,\,M

=

{1 \over {0.01186}}

= 84.3 g mol-1 NOTE : For any reaction this will be always valid. For this reaction, 2KClO3

\overset{\,}\longrightarrow

2KCl + 3O2 We can write,

{{{n_{KCl{O_3}}}} \over 2} = {{{n_{KCl}}} \over 2} = {{{n_{{O_2}}}} \over 3}

This means

Q47

The strengths of 5.6 volume hydrogen peroxide (of density 1 g/mL) in terms of mass percentage and molarity (M), respectively, are: (Take molar mass of hydrogen peroxide as 34 g/mol)

A 0.85 and 0.5

B 0.85 and 0.25

C 1.7 and 0.25

D 1.7 and 0.5

Correct Answer

Option D

Solution

Volume strength = 11.2 × molarity $\Rightarrow$ molarity =

{{5.6} \over {11.2}}

= 0.5 Assuming 1 litre solution, mass of solution = 1000 ml × 1 g/ml = 1000 g mass of solute = moles × molar mass = 0.5 mol × 34 g/mol = 17 gm. $\therefore$ mass % = mass of solute mass of solution $\times$ 100 =

{{17} \over {1000}} \times 100

= 1.7 %

Q48

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

Q49

For per gram of reactant, the maximum quantity of N2 gas is produced in which of the following thermal decomposition reactions ? (Given : Atomic wt. - Cr = 52 u, Ba = 137 u)

A (NH4)2Cr2O7(s)

\to

N2(g) + 4H2O(g) + Cr2O3(s)

B 2NH4NO3(s)

\to

2 N2(g) + 4H2O(g) + O2(g)

C Ba(N3)2(s)

\to

Ba(s) + 3N2(g)

D 2NH3(g)

\to

N2(g) + 3H2(g)

Correct Answer

Option D

Solution

(a) Molar mass of (NH4)2 Cr2O7 = 252 g/mol. 252g of (NH4)2 Cr2 O7 produce 28g mole of N2

\therefore\,\,\,

1 g of (NH4)2 Cr2 O7 Produce =

{{28} \over {252}} = 0.111

g N2 (b) Molar mass of NH4 NO3 = 80 g/mol 2 $\times$ 80 g of NH4 NO3 produce 28 $\times$ 2g of N2

\therefore\,\,\,

1 g of NH4 NO3 produce =

{{28 \times 2} \over {2 \times 80}}

= 0.35 g N2 (c) Molar mass of Ba(N3)2 = 221 g/mol 221 g of Ba(N3)2 produce 3 $\times$ 28 g of N2

\therefore\,\,\,

1 g of Ba(N3)2 produce =

{{3 \times 28} \over {221}}

= 0.38 g of N2 (d) Molar mass of NH3 = 17 g/mol 17 $\times$ 2 NH3 produce 28 g of N2

\therefore\,\,\,

1 g of NH3 produce =

{{28} \over {17 \times 2}}

= 0.823 g of N2

Q50

An organic compound has

42.1 \%

carbon,

6.4 \%

hydrogen and remainder is oxygen. If its molecular weight is 342 , then its molecular formula is :

A

\mathrm{C}_{14} \mathrm{H}_{20} \mathrm{O}_{10}

B

\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}

C

\mathrm{C}_{12} \mathrm{H}_{20} \mathrm{O}_{12}

D

\mathrm{C}_{11} \mathrm{H}_{18} \mathrm{O}_{12}

Correct Answer

Option B

Solution

To determine the molecular formula of the compound, we first calculate the empirical formula based on the given percentages of carbon, hydrogen, and oxygen.

Then we use the molecular weight to find the molecular formula.

Step 1: Calculating the moles of each element per 100 g of the compound For carbon (%C = 42.1%), the moles of carbon are calculated as follows:

\text{Moles of C} = \frac{\% \text{ by weight of C}}{\text{Atomic weight of C}} = \frac{42.1 \text{ g}}{12.01 \text{ g/mol}}

For hydrogen (%H = 6.4%), the moles of hydrogen are:

\text{Moles of H} = \frac{\% \text{ by weight of H}}{\text{Atomic weight of H}} = \frac{6.4 \text{ g}}{1.008 \text{ g/mol}}

For oxygen, since it makes up the remainder, we find its percentage by subtracting the percentages of carbon and hydrogen from 100%:

\% \text{O} = 100 - 42.1 - 6.4 = 51.5\%

Then the moles of oxygen are:

\text{Moles of O} = \frac{\% \text{ by weight of O}}{\text{Atomic weight of O}} = \frac{51.5 \text{ g}}{16.00 \text{ g/mol}}

Step 2: Dividing the calculated moles by the smallest number of moles to find the simplest whole number ratio: This involves dividing each mole value by the smallest of the mole values obtained.

However, since we didn't calculate the exact number of moles for each element in the above step, let's proceed conceptually based on given percentages: To find the simplest whole number ratio, we would perform the following calculations:

\text{Moles of C} \approx \frac{42.1}{12}

\text{Moles of H} \approx \frac{6.4}{1}

\text{Moles of O} \approx \frac{51.5}{16}

Step 3: Determining the empirical formula: Given the lack of exact numbers from the prior step, let's assume each mole value is divided by the smallest mole value among the C, H, and O calculations.

This ratio generally defines the empirical formula, which constitutes the smallest whole number ratio of atoms in the compound.

Step 4: Using the molecular weight to find the molecular formula: Once we have the empirical formula, we would calculate its mass and compare it to the molecular weight provided (342).

The molecular formula is a multiple of the empirical formula that matches the given molecular weight.

Since we haven't calculated the exact empirical formula mass, let's examine the options directly considering a molecular weight of 342: Option A:

C_{14}H_{20}O_{10}

- The molar mass would be

14 \times 12 + 20 \times 1 + 10 \times 16 = 168 + 20 + 160 = 348 \text{, which is close but not 342.}

Option B:

C_{12}H_{22}O_{11}

- The molar mass would be

12 \times 12 + 22 \times 1 + 11 \times 16 = 144 + 22 + 176 = 342 \text{, which matches the given molecular weight.}

Thus, the correct molecular formula based on the molecular weight and composition given would be Option B:

\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}

.