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8 months 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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8 months 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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8 months 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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8 months 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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8 months 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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8 months 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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8 months ago

Chemistry Class 12th

At 310 K, the solubility of CaF₂ in water is 2.34 * 10^-3 g/ 100 mL. The solubility product of CaF₂ is ____________* 10^-8 (mol/L)³. (Give molar mass : CaF₂ = 78 gmol^-1)

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Payal Gupta

Contributor-Level 10

Solubility of CaF₂ = S mol/L

$S = \frac{2.34 * 1 0^{- 3}}{0.1 * 78} = \frac{2.34}{78} * 1 0^{- 2} = 3 * 1 0^{- 4} m o l / L$

$K_{s p} (C a F_{2}) = 4 S^{3} = 4 {(3 * 1 0^{- 4})}^{3} = 108 * 1 0^{- 12}$

= 0.0108 * 10^-8 (mol/L)³

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8 months ago

Chemistry Class 11th

According to MO theory, number of species / ions from the following having identical bond order is___________.

CN^-, NO⁺, O₂, $O_{2}^{+}, O_{2}^{2 +}$

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Payal Gupta

Contributor-Level 10

$C N^{-}, N O^{+} a n d O_{2}^{2 +}$ have bond order = 3

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8 months ago

Chemistry Class 11th

The molar heat capacity for an ideal gas at constant pressure is 20.785JK^-1 mol^-1. The change in internal energy is 5000 J upon heating it from 300 K to 500 K. The number of moles of the gas at constant volume is __________. (Nearest Integer)

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Payal Gupta

Contributor-Level 10

$C_{P, m} = C_{v, m} + R$

$C_{v, m} = 20.785 - 8.314 = 12.471 J K^{- 1} m o l^{- 1}$

$n = \frac{5000}{12.471 * 200} = \frac{25}{12.471} \approx 2$

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8 months ago

Chemistry Class 11th

Amongst the following, the number of oxide(s) which are paramagnetic in nature is Na₂O, KO₂, NO₂, N₂O, ClO₂, NO, SO₂, Cl₂O (Nearest Integer)

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Payal Gupta

Contributor-Level 10

$K O_{2}, N O_{2}, C l O_{2}, N O$ are paramagnetic.

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