Ncert Solutions Chemistry Class 11th
Get insights from 2k questions on Ncert Solutions Chemistry Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Ncert Solutions Chemistry Class 11th
- 2k
Questions
- 0
Discussions
- 0
Active Users
- 29
Followers
Have a question about Ncert Solutions Chemistry Class 11th? Ask our experts
Keep it short & simple. Type complete word. Avoid abusive language.
Taking stability as the factor, which one of the following represents correct relationship?
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Taking stability as the factor, which one of the following represents correct relationship?
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Tl? ³ is less stable than Tl? ¹ (inert pair effect). Going down the group 13, stability of lower oxidation state increases. In case of B, Al, Ga and In, higher O.S. +3 remains more stable than lower O.S. +1 . But, in last stable element, thallium (Tl), lower O.S. +1 become more stable than higher O.
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R:
Assertion A: Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic.
Reason R: The deep blue solution is due to the formation of amide.
In the light of the above statements, choose the correct answer from the options given below:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R:
Assertion A: Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic.
Reason R: The deep blue solution is due to the formation of amide.
In the light of the above statements, choose the correct answer from the options given below:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
The para-magnetism of solution of alkali metals in liquid ammonia and its deep blue color is due to ammoniated electrons. Which absorb energy in visible region of light.
Which one of the following statements is correct?
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Which one of the following statements is correct?
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Ca plays important role in Neuromuscular function and interneuronal transmission. The daily requirement of Mg and Ca in the human body is estimated to be 200 – 300mg. All enzymes that utilise ATP in phosphate transfer require Mg as the cofactor. The bone in human body is NOT an inert and unchanging
Which of the following statements are NOT correct?
A. Hydrogen is used to reduce heavy metal oxides to metals.
B. Heavy water is used to study reaction mechanism.
C. Hydrogen is used to make saturated fats from oils.
D. The H – H bond dissociation enthalpy is lowest as compared to a single bond between two atoms of any element.
E. Hydrogen reduces oxides of metals that are more active than iron.
Choose the most appropriate answer from the options given below:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Which of the following statements are NOT correct?
A. Hydrogen is used to reduce heavy metal oxides to metals.
B. Heavy water is used to study reaction mechanism.
C. Hydrogen is used to make saturated fats from oils.
D. The H – H bond dissociation enthalpy is lowest as compared to a single bond between two atoms of any element.
E. Hydrogen reduces oxides of metals that are more active than iron.
Choose the most appropriate answer from the options given below:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
A, B and C are correct statements. The H-H bond dissociation enthalpy is the highest for a single bond between two atoms of any element. Hydrogen does not reduce oxides of metals that are more active than iron.
Which amongst the following options is correct graphical representation of Boyle's Law?
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Which amongst the following options is correct graphical representation of Boyle's Law?
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
PV = nRT
P = nRT . (1/V)
Plot of P vs (1/V) would be straight line passing through origin having slope = nRT.
At high temperatures, P vs (1/V) would have greater slope.
Intermolecular forces are forces of attraction and repulsion between interacting particles that will include:
A. dipole-dipole forces.
B. dipole - induced dipole forces.
C. hydrogen bonding.
D. covalent bonding.
E. dispersion forces.
Choose the most appropriate answer from the options given below:
<!--[endif]-->
Intermolecular forces are forces of attraction and repulsion between interacting particles that will include:
A. dipole-dipole forces.
B. dipole - induced dipole forces.
C. hydrogen bonding.
D. covalent bonding.
E. dispersion forces.
Choose the most appropriate answer from the options given below:
<!--[endif]-->
Covalent bonding is NOT an intermolecular force while rest all are considered as intermolecular forces.
The right option for the mass of CO? produced by heating 20 g of 20% pure limestone is (Atomic mass of Ca = 40)
[CaCO3 --(1200K)--> CaO + CO2]
The right option for the mass of CO? produced by heating 20 g of 20% pure limestone is (Atomic mass of Ca = 40)
[CaCO3 --(1200K)--> CaO + CO2]
20 g of 20% CaCO? = 20 × (20/100) = 4 g CaCO?
CaCO? (100 g) - (1200 K)-> CaO + CO? (44 g)
According to the reaction,
100 g of CaCO? → 44 g of CO?
4 g CaCO? → (4×44)/100 g of CO?
= 1.76 g of CO?
The slope of Arrhenius Plot (ln k v/s 1/T) of first order reaction is −5 × 103 K. The value of E? of the reaction is. Choose the correct option for your answer.
[Given R = 8.314JK-1 mol-1 ]
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
The slope of Arrhenius Plot (ln k v/s 1/T) of first order reaction is −5 × 103 K. The value of E? of the reaction is. Choose the correct option for your answer.
[Given R = 8.314JK-1 mol-1 ]
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
ln K = ln A – (E? /R) (1/T)
In ln K v/s 1/T graph
Slope = –E? /R
–5 × 10³ = –E? /8.314
E? = 5 × 10³ × 8.314
= 41500 J mol? ¹ or 41.5 kJ mol? ¹
The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in molar ratio 3:2 is:
[At 45°C vapour pressure of benzene is 280 mmHg and that of octane is 420 mmHg. Assume Ideal gas]
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in molar ratio 3:2 is:
[At 45°C vapour pressure of benzene is 280 mmHg and that of octane is 420 mmHg. Assume Ideal gas]
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
n (B) = 3/2
n (B) = 3, n (O) = 2
n (Total) = 3 + 2 = 5
X (B) = n (B) / n (T) = 3/5
X (O) = n (O) / n (T) = 2/5
P (S) = P° (B)X (B) + P° (O)X (O)
P (S) = 280 * (3/5) + 420 * (2/5)
= 336 mm of Hg
Dihedral angle of least stable conformer of ethane is:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Dihedral angle of least stable conformer of ethane is:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Dihedral angle (D.A.) of least stable conformer of ethane = 0°
D.A. = 0°
(Eclipsed)
