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Chemistry NCERT Exemplar Solutions Class 11th Chapter One

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Chemistry Coordination Compounds

Three moles of AgCl get precipitated when one mole of an octahedral co-ordination compound with empirical formula CrCl?·3NH?·3H?O reacts with excess of silver nitrate. The number of chloride ions satisfying the secondary valency of the metal ion is ______.

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Raj Pandey

Contributor-Level 9

CrCl? ·3NH? ·3H? O gives 3 moles of AgCl precipitate. This means all three Cl? are outside the coordination sphere.
The complex is [Cr (NH? )? (H? O)? ]Cl?
The 3 chloride ions satisfy only primary valency.
secondary valency satisfied by chloride ion = 0

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Ncert Solutions Chemistry Class 11th

A home owner uses 4.00 * 10³m³ of methane (CH?) gas, (assume CH? is an ideal gas) in a year to heat his home. Under the pressure of 1.0 atm and 300 K, mass of gas used is x * 10?g. The value of x is ______ (Nearest integer).
(Given R = 0.083 L atm K?¹mol?¹)

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Raj Pandey

Contributor-Level 9

PV = nRT
n = PV/RT = (1 atm * 4*10? L) / (0.083 LatmK? ¹mol? ¹ * 300 K) = 1.6 * 10? mol
Mass = n * Molar Mass = 1.6 * 10? mol * 16 g/mol = 25.6 * 10? g ≈ 26 * 10? g

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Ncert Solutions Chemistry Class 11th

A source of monochromatic radiation of wavelength 400 nm provides 1000 J of energy in 10 seconds. When this radiation falls on the surface of sodium, x * 10²? electrons are ejected per second. Assume that wavelength 400 nm is sufficient for ejection of electron from the surface of sodium metal. The value of x is ______ (Nearest integer).
(h = 6.626 * 10?³?Js)

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Raj Pandey

Contributor-Level 9

Energy per second = 1000 J / 10 s = 100 J/s
Energy of one photon E = hc/λ = (6.626*10? ³? * 3*10? ) / (400*10? ) = 4.965 * 10? ¹? J
Number of electrons ejected = Total energy / Energy per photon = 100 / (4.965 * 10? ¹? ) = 20.14 * 10¹? ≈ 2 * 10²?

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Class 11th

For the reaction A + B ? 2C
The value of equilibrium constant is 100 at 298K. If the initial concentration of all the three species is 1M each, then the equilibrium concentration of C is x * 10?¹M. The value of x is ______ (Nearest integer).

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Raj Pandey

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A + B? 2C
Initial: 1, 1
At eq: 1-x, 1+2x
K = [C]²/ ( [A] [B]) = (1+2x)²/ (1-x)² = 100
(1+2x)/ (1-x) = 10
1+2x = 10-10x => 12x = 9 => x = 3/4
[C] = 1+2x = 1+2 (3/4) = 1+1.5 = 2.5M = 25 * 10? ¹M

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Class 12th

Consider the cell at 25°C
Zn | Zn²?(aq), (1M) Fe³?(aq), Fe²?(aq) | Pt(s)
The fraction of total iron present as Fe³? ion at the cell potential of 1.500V is x * 10?². The value of x is ______ (Nearest integer).
(Given: E?(Fe³?/Fe²?) = 0.77V, E?(Zn²?/Zn) = -0.76V)

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Raj Pandey

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Fe? ³ + e? → Fe? ² E° = 0.77V
Zn (s) → Zn? ² + 2e? ; E° = 0.76V
Cell reaction: 2Fe? ³ + Zn → 2Fe? ² + Zn? ² E°cell = 1.53V
Ecell = E°cell - (0.059/2)log ( [Zn? ²] [Fe? ²]²/ [Fe? ³]²)
1.5 = 1.53 - (0.06/2)log (1 * [Fe? ²]²/ [Fe? ³]²)
-0.03 = -0.03 log ( [Fe? ²]/ [Fe? ³])²
1 = log ( [Fe? ²]/ [Fe? ³])² => [Fe? ²]/ [Fe? ³] = 10
Let total iron = T. [Fe? ³] + [Fe? ²] = T. [Fe? ³] + 10 [Fe? ³] = T. 11 [Fe? ³] = T.
fraction of Fe? ³ = [Fe? ³]/T = 1/11 ≈ 0.09
This solution seems to differ from the image. Let's follow the image's steps.
log ( [Fe? ²]/ [Fe? ³])² = 1 => ( [Fe? ²]/ [Fe? ³])² = 10
[Fe? ²]/ [Fe?

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Ncert Solutions Chemistry Class 12th

When 10 ml of an aqueous solution of Fe²? ions was titrated in the presence of dil H?SO? using diphenylamine indicator, 15 mL of 0.02 M solution of K?Cr?O? was required to get the end point. The molarity of the solution containing Fe²? ions is x * 10?²M. The value of x is ______. (Nearest integer)

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Raj Pandey

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Cr? O? ²? + 6Fe²? + 14H? → 2Cr³? + 6Fe³? + 7H? O
n-factor for Cr? O? ²? = 6, for Fe²? = 1
M.E Cr? O? ²? = M.E Fe²?
M? V? n? = M? V? n?
0.02 * 15 * 6 = M? * 10 * 1
M? = (0.02 * 15 * 6)/10 = 0.18 M = 18 * 10? ² M

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Chemistry Solutions

CO? gas is bubbled through water during a soft drink manufacturing process at 298 K. If CO? exerts a partial pressure of 0.835 bar then x m mol of CO? would dissolve in 0.9 L of water. The value of x is ______ (Nearest integer). (Henry's law constant for CO? at 298 K is 1.67*10³ bar)

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Raj Pandey

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P (CO? ) = K? X (CO? )
X (CO? ) = P (CO? )/K? = 0.835 / (1.67 * 10³) = 0.5 * 10? ³
X (CO? ) = n (CO? )/ (n (CO? ) + n (H? O) ≈ n (CO? )/n (H? O) (since n (CO? ) << n (H? O)
n (H? O) in 0.9L = 900g/18gmol? ¹ = 50 mol
n (CO? ) = X (CO? ) * n (H? O) = 0.5 * 10? ³ * 50 = 25 * 10? ³ moles = 25 mmol

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Ncert Solutions Chemistry Class 11th

Consider the complete combustion of butane, the amount of butane utilized to produce 72.0 g of water is ______ * 10?¹g, (in nearest integer).

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Raj Pandey

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C? H? + 13/2 O? → 4CO? + 5H? O
1 mole C? H? (58 g) produces 5 mole H? O (90 g)
∴ 90 g H? O obtained from 58 g C? H?
∴ 72g H? O obtained from (58/90) * 72g = 46.4 g
= 464 * 10? ¹g

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Ncert Solutions Chemistry Class 11th

At 298 K, the enthalpy of fusion of a solid (X) is 2.8kJ mol?¹ and the enthalpy of vaporization of the liquid (X) is 98.2 kJ mol?¹. The enthalpy of sublimation of the substance (X) in kJ mol?¹ is ______ (in nearest integer).

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