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Chemistry Chemical Kinetics

[A] -> [B]

Reactant Product

If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is x * 10^-⁶ s^-¹. The value of x is___________. (Nearest Integer)

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Vishal Baghel

Contributor-Level 10

$K = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{70 * 60} = \frac{6930}{7 * 6} * 1 0^{- 6}$

= 165 * 10^-6 s^-1

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a year ago

Chemistry Solutions

When a certain amount of solid A is dissolved in 100g of water at 25°C to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is 23.76 mmHg. The number of moles of solute A added is_________.

(Nearest Integer)

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Vishal Baghel

Contributor-Level 10

$\frac{P^{0} - P_{S}}{P^{0}} = \frac{n_{s o l u t e}}{n_{s o l v e n t}}$

$\frac{P^{0} - \frac{P^{0}}{2}}{P^{0}} = \frac{n_{s o l u t e}}{n_{s o l v e n t}}$

$n_{s o l u t e} = \frac{n_{s o l v e n t}}{2} = \frac{100}{18 * 2} = 2.78 m o l$

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a year ago

Chemistry Thermodynamics

A gas (Molar mass = 280 g mol^-¹) was burnt in excess O₂ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from 298.0 K to 298.45K. If the heat capacity of calorimeter is 2.5 kJ K^-¹ and enthalpy of combustion of gas is 9 kJ mol^-¹ the amount of gas burnt is _____________g. (Nearest Integer)

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Vishal Baghel

Contributor-Level 10

Let x gm is burnt

Moles = x/280

Heat released by x/280 mol = 2.5 * 0.45 kJ

Heat released by 1 mol = $\frac{2.5 * 0.45 * 280}{x} = k J$

$\begin{array}{l} Δ H = Δ U + Δ n g R T \\ Δ H = Δ U \end{array}$

$9 = \frac{2.5 * 280 * 0.45}{x}$

x = 35 gm

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a year ago

Chemistry Class 11th

For a real at 25°C temperature and high pressure (99 bar) the value of compressibility factor is 2, so the value of Vander Waal's constant 'b' should be__________* 10^-² L mol^-¹. (Nearest Integer)

(Given R = 0.083 L bar K^-¹ mol^-¹)

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Vishal Baghel

Contributor-Level 10

For real gas under high pressure,

$Z = 1 + \frac{P b}{R T} \Rightarrow b = \frac{R T}{P} = \frac{0.083 * 298}{99}$

= 0.25 * 10^-2 L mol^-1

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a year ago

Chemistry Chemistry Some Basic Concepts of Chemistry

The normality of H₂SO₄ in the solution obtained on mixing 100 mL of 0.1 M H₂SO₄ with 50 mL of 0.1 M NaOH is__________* 10^-¹N. (Nearest Integer)

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Vishal Baghel

Contributor-Level 10

No. of equivalents of H₂SO₄ = 100 * 0.1 * 2 = 20

No. of equivalents of NaOH = 50 * 0.1 = 5

No. equivalents of H₂SO₄ left = 20 – 5 = 15

150 * x = 15

$x = \frac{1}{10} = 0.1 N = 1 * 1 0^{- 1} N$

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a year ago

Chemistry Chemical Bonding and Molecular Structure

Why does MgO have a higher melting point than NaCl ?

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nitesh singh

Contributor-Level 10

Higher melting point means more energy is required to break the ionic bond. Since the ionic bonds in MgO are stronger than in NaCl due to various reasons, It have higher melting and even boiling point. Reasons are listed below:

In MgO, Magnesium and oxygen ions carry a +2 and -2 charge while sodium and in NaCl, chlorine ions carry a +1 and -1 charge. Due to higher charges more electrostatic attraction is involved.
The smaller ionic radius of Mg2+ are smaller compared to Na+, makes the bond stronger in MgO.
Due to larger elctrostatic force and smaller ionic radius, ionic bond in MgO are stronger, resulting in higher lattice energy.
The stro

...more

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a year ago

Chemistry Class 12th

In the following reaction

The % yield for reaction I is 60% and that of reaction II is 50%. The overall yield of the complete reaction is___________% [Nearest integer]

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Payal Gupta

Contributor-Level 10

Let initial moles of reactant taken = n. Total moles obtained for benzene sulphonic acid = 0.6 n (with % yield = 60%)

Moles of benzene sulphonic acid before reaction II = 0.6n

Moles obtained for phenol (with % yield 50%) = 0.6 * 0.5 n = 0.3 n

So, overall % age yield of complete reaction = $\frac{0.3 n}{n} * 100 = 30 %$

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New answer posted

a year ago

Chemistry Class 12th

Optical activity of an enantiomeric mixture is + 12.6° and the specific rotation of (+) isomer is +30°. The optical purity is_____________%

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Payal Gupta

Contributor-Level 10

$% O p t i c a l p u r i t y = \frac{O b s e r v e d r o t a t i o n o f m i x t u r e * 100}{r o t a t i o n o f p u r e e n a u t i o m e r}$

$= \frac{+ 12.6}{+ 30} * 100 = 42 %$

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a year ago

Chemistry Class 12th

In the titration of KMnO₄ and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is____________.

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Payal Gupta

Contributor-Level 10

$2 K M n O_{4} + 5 H_{2} {\overset{+ 3}{C}}_{2} O_{4} + 3 H_{2} S O_{4} (d i l) \to K_{2} S O_{4} + 2 M n S O_{4} + 10 C {\overset{+ 4}{O}}_{2} + 8 H_{2} O$

Oxidation state of carbon changes from +3 to +4

$∴$ change in O. S is 1

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a year ago

Chemistry Chemistry Coordination Compounds

The conductivity of a solution of complex with formula CoCl₃(NH₃)₄ corresponds to 1 : 1 electrolyte, then the primary valency of central metal ion is__________________.

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Payal Gupta

Contributor-Level 10

Complex is $[C o {(N H_{3})}_{4} C l_{2}] C l$

Primary valency = No. of ionisable species = 1

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