NEET Chemistry: Topic-wise Test- 3 Free Online Test 2026

Given, pH of Ba(OH)₂ = 12
pOH = 14-pH
= 14-12 = 2
We know that,
pOH = -log [OH^-]
2 =-log [OH^-]
[OH^-] = antilog (-2)
[OH^-] = 1 x 10^-2

Ba(OH)₂dissolves in water as 

Option D is correct.

Molar mass of Ca(OH)₂ = 40.08 + 2×(16.00 + 1.008) = 74.09 g mol^-1.

Moles of Ca(OH)₂ = 0.300 g ÷ 74.09 g mol^-1 = 0.00405 mol.

Molarity of Ca(OH)₂ = 0.00405 mol ÷ 0.500 L = 0.00810 M.

Each formula unit of Ca(OH)₂ produces two OH^- ions, so [OH^-] = 2 × 0.00810 = 0.01620 M.

pOH = -log[OH^-] = -log(0.01620) ≈ 1.790.

pH = 14.00 - 1.790 = 12.210, so the correct option is D (12.210).

∆H_reaction = (∆H_Product - ∆H_reactant)
= (-635.1 - 393.5)-(-1206.9) = 178.3 kJ mol^-1
So, we can say that the reaction is endothermic in nature. 
So, an increase in temperature will drive the reaction in a forward direction. (According to Le Chetelier principle-a principle stating that if a constraint (such as a change in pressure, temperature, or concentration of a reactant) is applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the constraint.)

PαX_solute (Henrys Law)

Work done =- ∫_Vi^Vf p_ext dVwhere V_i is initial and Vf is final volume. P_ex is the external pressure applied on system while change in volume

-ve sign is used as final volume v_f will be less than initial volume v_i  (v_f<v_i)
so to make final answer +ve a -ve sign is used .

Spontaneity means a reaction occurring on its own without any help of external agency.

when 1 mole of CO₂ is produced energy released is –393.5 kJ mol⁻¹ Moles of CO₂ given =35.2/44 =0.8 moles So energy released = 0.8 x393.5 KJ/mol = 315 KJ/mol

Specific heat will remain constant.

In case 1, the bond is broken in oxygen’s favor and it will attain its octet. Also, carbon becomes sp² hybridized, so there is a chance of polarity.
In case 2, if the bond is broken in favor of oxygen, then the ring will become anti-aromatic which is highly unstable and the bond won’t be broken in that way. If the bond is broken in favor of carbon in the ring, then although the ring becomes aromatic but oxygen will bear +ve charge which is very unstable. So, there is no chance to break the bond. 
In case 3, if the double bond is broken in favor of oxygen, then oxygen will acquire a negative charge and the ring will become aromatic. So, it is a highly favorable case of double bond breaking.
Therefore, the order of polarity: - III>I>II

The correct answer is Option D.
In 1,3,5-cyclohexatriene, there are three C = C having a bond length 134 pm and three C - C having a bond length 154 pm. In benzene, all the six C - C bonds have the same bond length 139 pm. 
 

The correct answer is Option D.

1, 3, 5, 7-cyclooctatetraene is a system with 8 electrons but it is a nonaromatic compound as it adopts a tub like shape to escape anti-aromaticity. Hence, d option is wrong.

• Chain isomerism occurs due to the possibility of branching in carbon chains.
• Geometrical isomerism or configurational isomerism or cis-trans isomerism arises when the functional group are present on same side of carbon or are on opposite side of the carbon chain.
• Metamerism arises when organic compounds have the same molecular formula but different alkyl groups on either side of the functional group.
• In C₂H₅OC₂H₅ and CH₃OCH₂CH₂CH₃  , we can see that in diethyl ether we have 2 carbon atoms on the left side of oxygen but in methyl propyl ether we have one carbon atom on the left of oxygen. On the right hand side of oxygen, we have 2 carbon atoms but in methyl propyl ether, we have 3 carbon atoms.

So, we can say that above molecules are metamers. Therefore, option C is correct.

4 aromatic isomers are possible.

For positional isomerism in alkene 4C-atom required →

For geometrical isomerism in alkene also 4c–atom required →

• Partition chromatography because the substances are partitioned or distributed between liquid phases. 
• The two phases are water held in pores of the filter paper and the other phase is a mobile phase that passes through the paper.

Alkenes like 1-hexene when flipped from top to bottom they have identical structures and also they have C=CH₂ unit which does not exist as cis-trans isomers.

Option D is correct.

For structure (i), identify the longest continuous carbon chain: it contains 5 carbon atoms, so the parent hydrocarbon is pentane.

Number the carbon chain from the end nearer the substituent so the substituent receives the lowest possible locant. A methyl substituent lies on carbon 3, giving the correct IUPAC name 3-methylpentane.

For structure (ii), the longest continuous carbon chain contains 4 carbon atoms, so the parent hydrocarbon is butane.

Number the chain to assign the lowest locant to the substituent. The methyl group is on carbon 2, giving the correct IUPAC name 2-methylbutane.

Therefore the pair of IUPAC names is 3-methylpentane and 2-methylbutane, matching option D.

To determine the IUPAC name of the given compound, follow these steps:

1. Identify the longest carbon chain: The longest continuous chain has 8 carbon atoms, so the base name is octane.
2. Number the chain: Number the carbon chain from the end nearest to the first substituent to give the substituents the lowest possible numbers.
3. Identify substituents and their positions:
   • At carbon 3, there is a methyl (-CH₃) group.
   • At carbon 4, there is an ethyl (-CH₂CH₃) group.
4. Write the name: Substituents are listed alphabetically regardless of their position numbers.
   • Ethyl comes before methyl alphabetically.
   • Therefore, the name is 4-ethyl-3-methyloctane.

Hence, the correct IUPAC name of the compound is 4-ethyl-3-methyloctane.

Answer: D. 4-ethyl-3-methyloctane

1° - 5, 2° - 1, 3° - 1, 4° - 1

Answer: (b), The IUPAC name of aforementioned compound is 1,1 –dimethyl -3-cyclohexanol.

H+  H⁺ + F^–
pK_w = pK_a + pKb
[For conjugate Acid-Base]
⇒ pK_a = 14 – 10.87 = 3.17
K_a = 6.76 × 10^–4

m. equivalent of KOH = 8
m. equivalent of HCOOH = 16
Remaining m. eq. (HCOOH) = 8
Formed m. eq. (HCOOK) = 8
⇒ Acidic Buffer
pH = pKa = 4 – log2
= 3.7
pOH = 10.3

NaX + HCl → NaCl + HX
[HX] at equivalent point = 0.05

Range of Indicator = 3 to 5

∆S = nCplnT₂/T₁ + nRlnP₁/P₂
Since pressure is constant, so the second term will be zero.
Or ∆S = 3×7/2×8.314×2.303×log(1000/400)
= 80.42 JK^-1

The reaction is 
C(diamond)     →     C(graphite)      ∆H = (94-91) = 3 kcal mol-1
∆S = ∆H/T
∆H = (94-91)×2.4×103/12   
= 600 kcal
∆S = 600/298 = 2.013 kcal K^-1

∆S_X→Z = ∆S_X→Y + ∆S_Y→Z(Entropy is a state function, so it is additive)
W_X→Y→Z = WX→Y (work done in _y→z is zero as the process is isochoric)

Volume in (I) = nRT/P₁ = 1× 0.0821× 298/4 = 6.11 L
Volume in (II) = nRT/P₂ = 1× 0.0821× 298/2 = 12.23 L
Total volume = 18.34 L
Applying PV = nRT at final condition, 
P = 2× 0.0821× 298/18.34 = 2.66