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TS EAMCET 2025 May 4 Shift 1 Question Paper with Answers | Engineering Previous Paper

131) 1 mL of ”x volume” $H_2O_2$ solution on heating gives 20 mL of oxygen gas at STP. The (w/v) % corresponding to ”x volume” of $H_2O_2$ is

A) 3.03

B) 6.06

C) 9.09

D) 30.3

View Answer

B) 6.06
Explanation:The decomposition reaction is $2H_2O_2 \rightarrow 2H_2O + O_2$ .
According to the stoichiometry, 2 moles of $H_2O_2$ ( $2 \times 34 = 68$ g) produce 22400 mL of $O_2$ at STP.
The problem states that 1 mL of the solution gives 20 mL of $O_2$ at STP.
Therefore, the volume strength of the $H_2O_2$ solution is 20 volumes.
Mass of $H_2O_2$ required to produce 22400 mL of $O_2$ is 68 g.
Mass of $H_2O_2$ required to produce 20 mL of $O_2$ is $\frac{68}{22400} \times 20$ g.
This calculated mass is the amount of $H_2O_2$ present in 1 mL of the solution.
To find the (w/v) %, we calculate the mass of $H_2O_2$ in 100 mL of the solution.
$(w/v) \% = \left( \frac{68 \times 20}{22400} \right) \times 100$ .
Simplifying the expression: $\frac{136000}{22400} = \frac{1360}{224} \approx 6.07$ .
This value corresponds closely to 6.06.

132) Identify the correct statements from the following
I. LiF is less soluble in water than NaF
II. Both LiCl and MgCl₂ are insoluble in ethanol
III. Both Li and Mg form nitrides
IV. Na₂CO₃ gives CO₂ on heating

A) I, IV

B) I, III

C) I, II

D) II, III

View Answer

B) I, III
Explanation:Mg-Li diagonal relation

133) The major ingredient (51%) in Portland cement is

A) Ca₂SiO₄

B) Ca₃SiO₅

C) Ca₃Al₂O₆

D) CaSO₄·2H₂O

View Answer

B) Ca₃SiO₅
Explanation:Main cement compound

134) Boron trifluoride on reaction with lithium aluminium hydride in ether gives LiF, AlF₃ and X. X on reaction with NH₃ gives Y. Y on further heating gives a compound Z. The number of σ-bonds and π-bonds in Z are x and y respectively. (x+y) is equal to

A) 15

B) 12

C) 14

D) 18

View Answer

A) 15
Explanation:Identification of X:
Boron trifluoride ( $BF_3$ ) reacts with lithium aluminium hydride ( $LiAlH_4$ ) in diethyl ether to produce Diborane ( $B_2H_6$ ).
$4BF_3 + 3LiAlH_4 \xrightarrow{ether} 2B_2H_6 + 3LiF + 3AlF_3$
Therefore, Compound X is Diborane ( $B_2H_6$ ).
Identification of Y and Z:
Diborane reacts with excess Ammonia ( $NH_3$ ) at low temperatures to form an ionic adduct, Borohydride-ammonia complex.
$B_2H_6 + 2NH_3 \rightarrow [BH_2(NH_3)_2]^+[BH_4]^-$
Therefore, Compound Y is $[BH_2(NH_3)_2][BH_4]$ .
On further heating this adduct to about 200°C, it decomposes to form Borazine ( $B_3N_3H_6$ ), also known as “Inorganic Benzene”.
$3[BH_2(NH_3)_2][BH_4] \xrightarrow{\Delta} 2B_3N_3H_6 + 12H_2$
Therefore, Compound Z is Borazine ( $B_3N_3H_6$ ).
Calculation of σ and π bonds in Z (Borazine):
Borazine has a hexagonal ring structure with alternating Boron (B) and Nitrogen (N) atoms.
– Each B atom is bonded to one H atom ( $\sigma$ -bond) and two N atoms in the ring.
– Each N atom is bonded to one H atom ( $\sigma$ -bond) and two B atoms in the ring.
σ-bonds (x):
– 6 $\sigma$ -bonds within the ring (B-N bonds).
– 3 $\sigma$ -bonds between Boron and Hydrogen (B-H).
– 3 $\sigma$ -bonds between Nitrogen and Hydrogen (N-H).
Total x = 6 + 3 + 3 = 12.
π-bonds (y):
– Similar to benzene, there are 3 double bonds (delocalized π-bonds) in the ring.
Total y = 3.
Final Calculation:
(x + y) = 12 + 3 = 15.

135) Consider the following sequence of reactions. 2CH₃Cl + Si $\xrightarrow[{573\;K}]{Cu}$ X $\xrightarrow{H_2O}$ Y $\xrightarrow{Polymerization}$ Z. The repeating structural unit in Z is

A) −[O−CH₃Si(O-)]−

B) −[(CH₃)₂Si-O]−

C) −[O−Si(CH₃)-O]− with another CH₃ attached to Si

D) −[CH₂ −Si(CH₃)-O]− with another CH₃ attached to Si

View Answer

B) −[(CH₃)₂Si-O]−
Explanation:Silicone polymer

136) Which of the following is not the common component of photochemical smog?

A) Ozone

B) Formaldehyde

C) Acrolein

D) SO2

View Answer

D) SO2
Explanation:SO2 not photochemical smog

137) Identify the compound (Z) in the following reaction sequence

A) Propanal

B) Propanone

C) Propanoic acid

D) Propanamide

View Answer

B) Propanone
Explanation:Reaction sequence ⇒ ketone

138) The correct IUPAC name of the following compound is

A) 5-Amino-4-hydroxy-1-bromo-1-cyanopentan-4-ol

B) 1-Amino-6-bromo-3-hydroxy-4-oxopentanenitrile

C) 6-Amino-2-bromo-5-hydroxy-3-oxohexanenitrile

D) 6-Amino-2-bromo-5-hydroxy-3-oxopentanenitrile

View Answer

C) 6-Amino-2-bromo-5-hydroxy-3-oxohexanenitrile
Explanation:-Identify the principal functional group: The cyanide (nitrile) group $-CN$ has the highest priority. It is assigned the number 1 carbon ( $C_1$ ).
– Determine the longest carbon chain: The chain starts from the nitrile carbon and includes all functional groups. In this compound, the chain consists of 6 carbons, making it a “hexanenitrile”.
– Number the chain:
$C_1$ : $-CN$ (Nitrile)
$C_2$ : $-Br$ (Bromo substituent)
$C_3$ : $=O$ (Oxo substituent/ketone)
$C_4$ : $-CH_2-$
$C_5$ : $-OH$ (Hydroxy substituent)
$C_6$ : $-NH_2$ (Amino substituent)
– Alphabetical order of substituents:
– 6-Amino
– 2-Bromo
– 5-Hydroxy
– 3-Oxo
– Combine the parts: 6-Amino-2-bromo-5-hydroxy-3-oxohexanenitrile.

139) Which purification method is generally used for a high boiling organic liquid compound, which decompose below its boiling point?

A) distillation

B) reduced pressure

C) steam distillation

D) fractional

View Answer

B) reduced pressure
Explanation:Vacuum distillation

140) Match compounds

List – 1 (Reaction)		List – 2 (Major product)
A	${\text{CH}}_3-\text{C}\equiv\text{CH}\xrightarrow[\underset{333k}{\mathrm{Hg}^{2+}/\text{H}^+}]{{\text{H}}_2O}$	I	${\text{CH}}_3-\underset{\overset\vert{\text{OH}}}{\text{CH}}-{\text{CH}}_3$
B	${\text{CH}}_3\text{COONa}\xrightarrow[{\;\mathrm{electrolysis}}]{{\text{H}}_2\text{O, Pt,}}$	II	${\text{CH}}_3-\underset{\overset{\vert\vert}{\text{O}}}{\text{C}}-{\text{CH}}_3$
C	$\text{CH}_3 - \text{CH} = \text{CH}_2 \xrightarrow{\text{H}_2\text{O}/\text{H}^+}$	III	${\text{CH}}_3-\underset{\overset\vert{\text{OH}}}{\text{CH}}-\underset{\overset\vert{\text{OH}}}{{\text{CH}}_2}$
D	${\text{CH}}_3-\text{CH}={\text{CH}}_2\xrightarrow[\underset{273\;K}{\mathrm{dil}.{\mathrm{KMnO}}_4}]{{\text{H}}_2O}$	IV	$\text{CH}_3 - \text{CH}_3$
		V	$\text{CH}_4$

A) A-II,B-IV,C-I,D-III

B) A-II,B-V,C-I,D-III

C) A-III,B-IV,C-II,D-I

D) A-II,B-III,C-IV,D-I

View Answer

A) A-II,B-IV,C-I,D-III
Explanation:A.CH₃–C≡CH + H₂O / Hg²⁺, H⁺
→ Markovnikov addition → enol → keto form
→ CH₃–CO–CH₃ (acetone)
⇒ A → II
B.CH₃COONa (electrolysis, Kolbe reaction)
→ decarboxylation → CH₃• radicals combine
→ CH₃–CH₃ (ethane)
⇒ B → IV
C.CH₃–CH=CH₂ + H₂O / H⁺
→ Markovnikov hydration
→ CH₃–CHOH–CH₃ (isopropanol)
⇒ C → I
D.CH₃–CH=CH₂ + dil. KMnO₄ (cold)
→ vicinal diol formation
→ CH₃–CHOH–CH₂OH
⇒ D → III
Final Matching:
A → II
B → IV
C → I
D → III

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