Some Basic Concepts of Chemistry — Page 4 | JEE Chemistry

Q31

5 moles of AB2 weigh 125 × 10–3 kg and 10 moles of A2B2 weigh 300 × 10–3 kg. The molar mass of A (MA) and molar mass of B (MB) in kg mol are:

A MA = 10 × 10–3 and MB = 5 × 10–3

B MA = 25 × 10–3 and MB = 50 × 10–3

C MA = 5 × 10–3 and MB = 10 × 10–3

D MA = 50 × 10–3 and MB = 25 × 10–3

Correct Answer

Option C

Solution

Molar weigh of 'A' = MA Molar weigh of 'B' = MB Molar mass of AB2 = MA + 2MB Molar mass of A2B2 = 2(MA + MB) 5 mol AB2 weighs 125 g $\therefore$ AB2 = 25 g/mol 10 mol A2B2 weighs 300 g $\therefore$ A2B2 = 30 g/mol So, MA + 2MB = 25 ..............(1) and 2(MA + MB) = 30 ..............(

2) By solving (1) and (2), MA = 5 gm = 5 $\times$ 10-3 kg MB = 10 gm = 10 $\times$ 10-3 kg

Q32

250 \mathrm{~g}

solution of

\mathrm{D}

-glucose in water contains

10.8 \%

of carbon by weight. The molality of the solution is nearest to (Given: Atomic Weights are,

\mathrm{H}, 1 \,\mathrm{u} ; \mathrm{C}, 12 \,\mathrm{u} ; \mathrm{O}, 16 \,\mathrm{u}

)

A 1.03

B 2.06

C 3.09

D 5.40

Correct Answer

Option B

Solution

Weight of D-glucose in water $=250 \mathrm{~g}$

\begin{aligned} \therefore \text{ Weight of carbon in D-glucose } &=\frac{250}{180} \times 72 \\\\ &=100 \mathrm{~g} \end{aligned}

$\%$ of carbon in the aqueous solution of glucose is $=10.8 \%$ $\therefore$ Weight of the solution is $=925.93$

\begin{aligned} \therefore \text{ Molality of D-glucose is } &=\frac{\frac{250}{180}}{(925.93-250)} \times 1000 \\\\ &=\frac{250}{180 \times 675.93} \times 1000 \\\\ &=2.06 \end{aligned}

Q33

\begin{aligned} &\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+400 \mathrm{~kJ} \\ &\mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+100 \mathrm{~kJ} \end{aligned}

When coal of purity 60% is allowed to burn in presence of insufficient oxygen, 60% of carbon is converted into 'CO' and the remaining is converted into '

\mathrm{CO}_{2}

'. The heat generated when

0.6 \mathrm{~kg}

of coal is burnt is _________.

A 1600 kJ

B 3200 kJ

C 4400 kJ

D 6600 kJ

Correct Answer

Option D

Solution

Weight of coal $=0.6 \mathrm{~kg}=600 \mathrm{gm}$ $\therefore 60 \%$ of it is carbon So weight of carbon $=600 \times \dfrac{60}{100}=360 \mathrm{~g}$ $\therefore$ moles of carbon $=\dfrac{360}{12}=30$ moles

\begin{gathered} \underset{12 \text{ moles }}{\mathrm{C}}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2} \\ \underset{\substack{18 \text{ moles } \\ \text{ (60\% of total Carbon })}}{\mathrm{C}}+\frac{1}{2} \mathrm{O}_{2} \longrightarrow \mathrm{CO} \end{gathered}

$\therefore$ Heat generated $=12 \times 400+18 \times 100=6600 \mathrm{~kJ}$

Q34

When a hydrocarbon A undergoes combustion in the presence of air, it requires 9.5 equivalents of oxygen and produces 3 equivalents of water. What is the molecular formula of A?

A

\mathrm{C_9H_6}

B

\mathrm{C_6H_6}

C

\mathrm{C_8H_6}

D

\mathrm{C_9H_9}

Correct Answer

Option C

Solution

{C_x}{H_y} + \left( {x + {y \over 4}} \right){O_2} \to xC{O_2} + {y \over 2}{H_2}O

{y \over 2} = 3

y = 6

x + {y \over 4} = {{19} \over 2}

x = {{19} \over 2} - {3 \over 2} = 8

So, formula is

{C_8}{H_6}

.

Q35

25 ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35 ml. The molarity of barium hydroxide solution was

A 0.14

B 0.28

C 0.35

D 0.07

Correct Answer

Option D

Solution

Ba(OH)2 + 2HCl(aq) $\to$ BaCl2(aq) + 2H2O(l) Assume molarity of Ba(OH)2 = M No of moles of Ba(OH)2 present = 25 $\times$ M No of moles of HCl present = 35 $\times$ 0.1 Stoichiometric coefficient of Ba(OH)2 = 1 Stoichiometric coefficient of HCl = 2 $\therefore$

{{25 \times M} \over 1}

=

{{35 \times 0.1} \over 2}

$\Rightarrow$ M = 0.07

Q36

1 \mathrm{~L}, 0.02 \mathrm{M}

solution of

\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right]

Br is mixed with

1 \mathrm{~L}, 0.02 \mathrm{M}

solution of

\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}

. The resulting solution is divided into two equal parts

(\mathrm{X})

and treated with excess of

\mathrm{AgNO}_{3}

solution and

\mathrm{BaCl}_{2}

solution respectively as shown below:

1 \mathrm{~L}

Solution

(\mathrm{X})+\mathrm{AgNO}_{3}

solution (excess)

\longrightarrow \mathrm{Y}

1 \mathrm{~L}

Solution

(\mathrm{X})+\mathrm{BaCl}_{2}

solution (excess)

\longrightarrow \mathrm{Z}

The number of moles of

\mathrm{Y}

and

\mathrm{Z}

respectively are

A

0 .01,0.01

B

0.01,0.02

C

0.02,0.01

D

0.02,0.02

Correct Answer

Option A

Solution

On mixing both

\mathrm{\left[ {Co{{(N{H_5})}_5}S{O_4}} \right]Br}

and

\mathrm{\left[ {Co{{(N{H_3})}_5}Br} \right]S{O_4}}

becomes 0.01 molar. $\therefore$ Moles of y and z formed will also be 0.01 both.

Q37

A solution is prepared by adding

2 \mathrm{~g}

of "

\mathrm{X}

" to 1 mole of water. Mass percent of "

\mathrm{X}

" in the solution is :

A 20%

B 10%

C 2%

D 5%

Correct Answer

Option B

Solution

First, we need to calculate the mass of 1 mole of water (H2O).

The molecular weight of water is 18 g/mol, so 1 mole of water weighs 18 g.

The total mass of the solution is the mass of "X" plus the mass of water, which is 2 g + 18 g = 20 g.

We can calculate the mass percent of "X" in the solution with the following formula:

\text{Mass percent of "X"} = \left(\frac{\text{mass of "X"}}{\text{total mass of solution}}\right) \times 100\%

Substitute the given values into the formula :

\text{Mass percent of "X"} = \left(\frac{2 \, \text{g}}{20 \, \text{g}}\right) \times 100\% = 10\%

Therefore, Option B, $10\%$ , is the correct answer.

Q38

The volume of

0.02 ~\mathrm{M}

aqueous

\mathrm{HBr}

required to neutralize

10.0 \mathrm{~mL}

of

0.01 ~\mathrm{M}

aqueous

\mathrm{Ba}(\mathrm{OH})_{2}

is (Assume complete neutralization)

A 7.5 mL

B 5.0 mL

C 10.0 mL

D 2.5 mL

Correct Answer

Option C

Solution

The reaction between $\mathrm{HBr}$ and $\mathrm{Ba(OH)_2}$ is a neutralization reaction where one $\mathrm{Ba(OH)_2}$ reacts with two $\mathrm{HBr}$ to form $\mathrm{BaBr_2}$ and $2\mathrm{H_2O}$ : $\mathrm{Ba(OH)_2} + 2\mathrm{HBr} \rightarrow \mathrm{BaBr_2} + 2\mathrm{H_2O}$ We can use the stoichiometry of the reaction to find the volume of $\mathrm{HBr}$ required.

The number of moles of $\mathrm{Ba(OH)_2}$ is given by its molarity times the volume in liters: $n(\mathrm{Ba(OH)_2}) = 0.01 \, \mathrm{M} \times \dfrac{10.0 \, \mathrm{mL}}{1000 \, \mathrm{mL/L}} = 0.0001 \, \mathrm{moles}$ From the balanced equation, we can see that the reaction requires $2$ moles of $\mathrm{HBr}$ for each mole of $\mathrm{Ba(OH)_2}$ : $n(\mathrm{HBr}) = 2 \times n(\mathrm{Ba(OH)_2}) = 2 \times 0.0001 \, \mathrm{moles} = 0.0002 \, \mathrm{moles}$ Then, we find the volume of the $\mathrm{HBr}$ solution by dividing the number of moles by the molarity: $V(\mathrm{HBr}) = \dfrac{n(\mathrm{HBr})}{0.02 \, \mathrm{M}} = \dfrac{0.0002 \, \mathrm{moles}}{0.02 \, \mathrm{mol/L}} = 0.01 \, \mathrm{L} = 10 \, \mathrm{mL}$ Therefore, the volume of the $0.02 \, \mathrm{M}$ $\mathrm{HBr}$ solution required to neutralize the $\mathrm{Ba(OH)_2}$ solution is $10.0 \, \mathrm{mL}$ .

Q39

The Molarity (M) of an aqueous solution containing

5.85 \mathrm{~g}

of

\mathrm{NaCl}

in

500 \mathrm{~mL}

water is : (Given : Molar Mass

\mathrm{Na}: 23

and

\mathrm{Cl}: 35.5 \mathrm{~gmol}^{-1}

)

A 20

B 4

C 2

D 0.2

Correct Answer

Option D

Solution

To find the molarity of the aqueous solution, we need to calculate the number of moles of NaCl and then divide it by the volume of the solution in liters.

First, determine the molar mass of NaCl:

\text{Molar mass of NaCl} = \text{Molar mass of Na} + \text{Molar mass of Cl} = 23 + 35.5 = 58.5 \, \text{g/mol}

Next, calculate the moles of NaCl:

\text{Moles of NaCl} = \frac{5.85 \, \text{g}}{58.5 \, \text{g/mol}} = 0.1 \, \text{mol}

Now, convert the volume of the solution from milliliters to liters:

500 \, \text{mL} = 0.500 \, \text{L}

Finally, calculate the molarity (M), which is the number of moles of solute per liter of solution:

M = \frac{\text{Moles of NaCl}}{\text{Volume of solution in liters}} = \frac{0.1 \, \text{mol}}{0.500 \, \text{L}} = 0.2 \, \text{M}

Therefore, the molarity of the solution is 0.2 M. The correct option is D.

Q40

Combustion of glucose

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

produces

\mathrm{CO}_2

and water. The amount of oxygen (in

\mathrm{g}

) required for the complete combustion of

900 \mathrm{~g}

of glucose is : [Molar mass of glucose in

\mathrm{g} \mathrm{~mol}^{-1}=180

]

A 32

B 480

C 800

D 960

Correct Answer

Option D

Solution

To determine the amount of oxygen required for the complete combustion of 900 g of glucose, we need to follow these steps: First, let's write the balanced chemical equation for the combustion of glucose

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

:

\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 + 6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2 + 6 \mathrm{H}_2 \mathrm{O}

From the balanced equation, we see that 1 mole of glucose reacts with 6 moles of oxygen

(\mathrm{O}_2)

. Next, let's calculate the number of moles of glucose in 900 g. The molar mass of glucose is given as

180 \ \mathrm{g} \ \mathrm{mol}^{-1}

. Number of moles of glucose =

\frac{900 \ \mathrm{g}}{180 \ \mathrm{g} \ \mathrm{mol}^{-1}} = 5 \ \mathrm{mol}

According to the balanced equation, 1 mole of glucose requires 6 moles of oxygen.

Therefore, 5 moles of glucose will require: Number of moles of oxygen =

5 \ \mathrm{mol} \times 6 = 30 \ \mathrm{mol}

Now, we need to find the mass of 30 moles of oxygen. The molar mass of oxygen

\left(\mathrm{O}_2\right)

is

32 \ \mathrm{g} \ \mathrm{mol}^{-1}

. Mass of oxygen =

30 \ \mathrm{mol} \times 32 \ \mathrm{g} \ \mathrm{mol}^{-1} = 960 \ \mathrm{g}

Therefore, the amount of oxygen required for the complete combustion of 900 g of glucose is 960 g.

So, the correct option is Option D: 960.