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At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O₂ by volume for complete combustion. After combustion, the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is :
Option: 1 C₄H₈
Option: 2 C₄H₁₀
Option: 3 C₃H₆
Option: 4 C₃H₈

Volume of N₂ in air = 375 × 0.8 = 300 ml

Volume of O₂ in air = 375 × 0.2 = 75 ml

$C_{x}H_{y} +\left ( x +\frac{y}{4} \right )O_{2} \; \rightarrow \; xCO_{2}(g) + \frac{y}{2} H_{2}O(l)$

15ml $15\left ( x +\frac{y}{4} \right )$

0 0 15x -

After combustion total volume

$330 =V_{N_{2}} + V_{CO_{2}}$

330 = 300 + 15x

x = 2

Volume of O₂used

$15\left ( x +\frac{y}{4} \right ) = 75$

$\left ( x +\frac{y}{4} \right ) = 5$

y = 12

So hydrocarbon is = C₂H₁₂

None of the options matches it therefore it is a BONUS.

----------------------------------------------------------------------

Alternatively Solution

$C_{x}H_{y} +\left ( x +\frac{y}{4} \right )O_{2} \; \rightarrow \; xCO_{2}(g) + \frac{y}{2} H_{2}O(l)$

15ml $15\left ( x +\frac{y}{4} \right )$

0 0 15x -

Volume of O₂ used

$15\left ( x +\frac{y}{4} \right ) = 75$

$\left ( x +\frac{y}{4} \right ) = 5$

If further information (i.e., 330 ml) is neglected, option (C₃H₈ ) only satisfy the above equation.

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Ritika Jonwal

An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVⁿ=constant, then n is given by (Here C_P and C_V are molar specific heat at constant pressure and constant volume, respectively)
Option: 1   $n=\frac{C_{p}}{C_{v}}$

Option: 2   $n=\frac{C-C_{p}}{C-C_{v}}$

Option: 3 $n=\frac{C_{p}-C}{C-C_{v}}$

Option: 4   $n=\frac{C-C_{v}}{C-C_{p}}$

For a polytropic preocess

$c=c_{v}+\frac{R}{1-n} \: or \: \frac{R}{1-n} = c-c_{v}$

$\Rightarrow 1-n=\frac{R}{c-c_{v}} \: or\: n=1-\frac{R}{c-c_{v}}$

$\Rightarrow n=\frac{c-\left ( c_{v}+R \right )}{c-c_{v}} = \frac{c-c_{p}}{c-c_{v}}$

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Ritika Jonwal

A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals µ. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction µ and the distance x(=QR), are, respectively close to :
Option: 1 0.2 and 6.5 m
Option: 3 0.2 and 3.5 m
Option: 4 0.29 and 6.5 m

Work done by friction at QR = μmgx

In triangle, sin 30° = 1/2 = 2/PQ

PQ = 4 m

Work done by friction at PQ = μmg × cos 30° × 4 = μmg × √3/2 × 4 = 2√3μmg

Since work done by friction on parts PQ and QR are equal,

μmgx = 2√3μmg

x = 2√3 ≅ 3.5 m

Applying work energy theorem from P to R

decrease in P.E.=P.E.= loss of energy due to friction in PQPQ and QR

$\\ m g h=(\mu m g \cos \theta) P Q+\mu m g \times Q R\\ h=\mu \cos \theta \times P Q+\mu m g \times Q R\\ h=\mu \cos \theta \times P Q+\mu \times Q R =\mu \cos 30^{\circ} \times 4+\mu \times 2 \sqrt{3} =\mu\left(4 \times \frac{\sqrt{3}}{2}+2 \sqrt{3}\right)\\ h=\mu \times 4 \sqrt{3}\\ \mu=\frac{2}{4 \sqrt{3}}=\frac{1}{2 \sqrt{3}}=0.29$ where h=2(given)

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Ritika Jonwal

In the following structure, the double bonds are marked as I, II, III and IV Geometrical isomerism is not possible at site (s) :
Option: 1 III
Option: 2 I
Option: 3 I and II
Option: 4 III and IV

Geometrical isomerism is not possible at Site I as two identical methyl groups are attached to the same carbon bearing the double bond.

Hence, the answer is Option (2)

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vishal kumar

Which one of the following is an oxide ?
Option: 1 KO₂
Option: 2 BaO₂
Option: 3 SiO₂
Option: 4 CsO₂

SiO₂ - oxide

KO_???2? , CsO_????2 - superoxides

BaO_????2 - peroxide

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vishal kumar

The following reaction occurs in the Blast Furnace where iron ore is reduced to iron metal : Fe₂O₃(s)+3 CO(g) 2 Fe(l)+3 CO₂(g) Using the Le Chatelier’s principle, predict which one of the following will not disturb the equilibrium ?
Option: 1 Removal of CO
Option: 2 Removal of CO₂
Option: 3 Addition of CO₂
Option: 4 Addition of Fe₂O₃

$\\Fe_{2}O_{3}+3CO \leftrightharpoons 2Fe+3CO \\$

According to Le Chatelier's principle change in concentration by changing the amount of reactant or product affect the equilibrium. However, the addition of solid reactant won't affect the concentration.

Therefore, addition of solid Fe₂O_???3 will not disturb the equilibrium.

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vishal kumar

The electronic configuration with the highest ionization enthalpy is :
Option: 1 [Ne] 3s² 3p¹
Option: 2 [Ne] 3s² 3p²
Option: 3 [Ne] 3s² 3p³
Option: 4 [Ar] 3d¹⁰ 4s² 4p³

The electronic configuration with the highest ionization enthalpy is [Ne]3s23p3. On moving down the group, the ionization enthalpy decreases. In a period, on moving from left tor fight, the ionization enthalpy increases.

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vishal kumar

A solution is prepared by mixing 8.5 g of CH₂Cl₂ and 11.95 g of CHCl₃. If vapour pressure of CH₂Cl₂ and CHCl₃ at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl₃ in vapour form is : (Molar mass of Cl=35.5 g mol⁻¹)
Option: 1 0.162
Option: 2 0.675
Option: 3 0.325
Option: 4 0.486

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vishal kumar

What quantity (in mL) of a 45% acid solution of a monoprotic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution?
Option: 1 320
Option: 2 325
Option: 3 316
Option: 4 330

Reactions in Solutions -

Concentration :
It is the amount of solute present in one litre of solution. It is denoted by C or S.

$\\ \mathrm{C\, or\, S\, =\, \frac{Weight\, of\, solute\, in\, gram}{Volume\, in\, litre}} \\\\ \mathrm{C\, =\, N\, \times\, E}\\\\\mathrm{Here\, N\, =\, normality\, and \, E\, =\, Eq.\, wt.}$

Mole Fraction:
It is the ratio of moles of one component to the total number of moles present In the solution. It is expressed by X for example, for a binary solution with two components A and B.

$X_{A}\, = \frac{n_{A}}{n_{A}\, +\, n_{B}}$

$X_{B}\, = \frac{n_{B}}{n_{A}\, +\, n_{B}}$

$X_{A}\, +\, X_{B}\, =\, 1$

Here n_A and n_B represent moles of solvent and solute respectively. Mole fraction does not depend upon temperature as both solute and solvent are expressed by weight.

Molarity:
It is the number of moles or gram moles of solute dissolved per litre of the solution. Molarity is denoted by 'M'.

$\mathrm{M\, =\, \frac{Weight\, of\, solute\, in\, gram}{Molar\, mass\, \times\, volume\, in\, litre}}$

When molarity of a solution is one, it is called a molar solution and when it is 0.1, solution is called decimolar solution.
Molarity depends upon temperature and its unit is mol/litre.
M₁V₁ = M₂V₂
On dilution water added = V₂ - V₁
When the solution of two different substances react together then
$\mathrm{\frac{M_{1}V_{1}}{n_{1}}\, =\, \frac{M_{2}V_{2}}{n_{2}}}$
Here M, V, n are molarity, volume and number of molecules taking part in a reaction respectively.
When a mixture of different solutions having different concentrations are taken the molarity of the mixture is calculated as follows:
$\mathrm{M\, =\, \frac{M_{1}V_{1}\, + M_{2}V_{2}\,.....}{V_{1}\, +\, V_{2}.....}}$
When density and % by weight of a substance in a solution are given, molarity is find as follows:
$\mathrm{M\, =\,\frac{\%\, by\, weight\,\times\, d\, \times\, 10}{Molecular\, weight}}$
Here d = density

Molality
It is the number of moles or gram moles of solute dissolved per kilogram of the solvent. It is denoted by 'm'.
$\mathrm{m\, =\, \frac{Weight\, of\, solute\, in\, gram }{Molar\, mass\, \times\, wt.\, of\, solvent\, in\, Kg}}$

If molality is one solution, it is called molal solution.
One molal solution is less than one molar solution.
Molality is preferred over molarity during experiments as molality is temperature independent while molarity is temperature-dependent.

Normality
It is the number of gram equivalents of solute present in one litre of the solution and it is denoted by 'N'.
$\mathrm{N\, =\, \frac{Weight\, of\, solute\, in\, gram }{Equivalent\, mass\, \times\, volume\, in\, litre }}$

When normality of a solution is one, the solution is called normal solution and when it is 0.1, the solution is called deci-normal solution.

Normality Equation:

$\mathrm{N_{1}V_{1}\, =\, N_{2}V_{2}}$

Volume Of water added = V₂ - V₁
Here V₂ = volume after dilution
V₁ = volume before dilution
When density and % by weight of a substance in a solution are given, normality is find as follows:

$\mathrm{N\, =\, \frac{\%\, by\, weight\, \times\, d \times\, 10 }{Equivalent\ weight }}$
Here d = density of solution
When a mixture of different solutions having different concentrations are taken the normality of the mixture is calculated as follows:
$\mathrm{N\, =\, \frac{N_{1}V_{1}\, + N_{2}V_{2}\,.....}{V_{1}\, +\, V_{2}.....}}$
• In case of acid-base neutralization the normality of the resulting solution
$\mathrm{N\, =\, \frac{N_{1}V_{1}\, - \, N_{2}V_{2}\,.....}{V_{1}\, +\, V_{2}.....}}$
• To find weight of substance
$W\, =\, \frac{NEV}{1000}$

Relation between Normality and Molarity :
N x Eq wt. = molarity x molar mass
N = molarity x valency
N = molarity x number of H⁺ or OH^- ion

-

As we have learnt,

$\\\mathrm{\frac{V\times45}{100}\,+\,\frac{(800-V)20}{100}\, =\, \frac{800\times29.875}{100}}\\\\\mathrm{\frac{9V}{20}\, +\, 160\, -\, \frac{V}{5}\, =\, 239}\\\\\mathrm{\frac{5V}{20}\, =\, 79\,}\\\\\mathrm{\therefore V\, =\, 316}$

Therefore, Option (3) is correct

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vishal kumar

At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen (N₂) at 4 bar. The molar mass of gaseous molecule is :
Option: 1 28 g mol⁻¹
Option: 2 56 g mol⁻¹
Option: 3 112 g mol⁻¹
Option: 4 224 g mol⁻¹

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In the following structure, the double bonds are marked as I, II, III and IV Geometrical isomerism is not possible at site (s) : Option: 1 III Option: 2 I Option: 3 I and II Option: 4 III and IV

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Which one of the following is an oxide ? Option: 1 KO2 Option: 2 BaO2 Option: 3 SiO2 Option: 4 CsO2

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vishal kumar

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In the following structure, the double bonds are marked as I, II, III and IV Geometrical isomerism is not possible at site (s) :
Option: 1 III
Option: 2 I
Option: 3 I and II
Option: 4 III and IV

Which one of the following is an oxide ?
Option: 1 KO₂
Option: 2 BaO₂
Option: 3 SiO₂
Option: 4 CsO₂

The electronic configuration with the highest ionization enthalpy is :
Option: 1 [Ne] 3s² 3p¹
Option: 2 [Ne] 3s² 3p²
Option: 3 [Ne] 3s² 3p³
Option: 4 [Ar] 3d¹⁰ 4s² 4p³

What quantity (in mL) of a 45% acid solution of a monoprotic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution?
Option: 1 320
Option: 2 325
Option: 3 316
Option: 4 330

At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen (N₂) at 4 bar. The molar mass of gaseous molecule is :
Option: 1 28 g mol⁻¹
Option: 2 56 g mol⁻¹
Option: 3 112 g mol⁻¹
Option: 4 224 g mol⁻¹