Chemistry Test 295

Result

Chemistry Test 295

Score
-
out of -
Rank
-
out of -

TIME Taken - -

Weekly Quiz Competition

Question 1
4 / -1

The hydrolysis of sucrose has an equilibrium constant Kc = 2 × 10¹³ at 300 K. What is the sign of ΔG^∘ for this reaction?

A
Positive

B
Negative

C
Zero

D
Cannot be determined

Solution

Since K > 1, ln K is positive, making ΔG^∘ negative. This indicates a spontaneous reaction under standard conditions.
Question 2
4 / -1

According to Kohlrausch’s law, the limiting molar conductivity of CaCl₂ can be expressed as:

A

B

C

D

Solution
Question 3
4 / -1

The experimentally observed bond length in the ozone (O₃) molecule is 128 pm. Which of the following best explains this observation?

A
The molecule rapidly interconverts between two structures with bond lengths of 121 pm and 148 pm.

B
The actual structure is an average of two resonance forms, resulting in a bond length intermediate between a single and double bond.

C
The molecule has one single bond and one double bond, but the bonds vibrate to give an average length.

D
The bonds in ozone are purely double bonds due to delocalization.

Solution

The bond length in O₃ (128 pm) lies between the typical O − O single bond (148 pm) and O = O double bond (121 pm). This is explained by resonance, where the true structure is a hybrid of two canonical forms, averaging the bond characteristics.
Question 4
4 / -1

A newly discovered co-enzyme C required for the activity of a human oxidoreductase is suspected to be either a nicotinamide-derived vitamin or a riboflavin-derived vitamin. A 0.50 mM aqueous solution of C shows an absorbance of 0.42 at 260 nm in a 1.0 cm cuvette, but no significant absorbance at 450 nm. When 1.0 × 10⁻⁴ mol of C is completely hydrolysed, it yields 1.0 × 10⁻⁴ mol of a phosphoric acid–stable fragment that gives a positive Tollens’ test and reduces Benedict’s solution at pH 10.7, 298 K. The same molar quantity of C, upon exhaustive enzymatic dephosphorylation followed by acid hydrolysis, gives 2.0 × 10⁻⁴ mol of NH₃ and 1.0 × 10⁻⁴ mol of a neutral, optically active, 6-carbon aldaric acid whose meso-form is not optically active. Identify the correct statement about C.

A
It is NADP⁺ and the 6-carbon aldaric acid is D-glucaric acid.

B
It is FAD and the 6-carbon aldaric acid is L-idaric acid.

C
It is NAD⁺ and the 6-carbon aldaric acid is D-mannaric acid.

D
It is NADP⁺ and the 6-carbon aldaric acid is L-gularic acid.

Solution

Step 1: Absence of absorbance at 450 nm rules out the yellow iso-alloxazine system of riboflavin-derived co-enzymes (FAD, FMN). The absorbance at 260 nm (A = 0.42, path-length 1 cm) gives

which matches the adenine chromophore present in nicotinamide nucleotides (NAD⁺/NADP⁺). Hence the co-enzyme is nicotinamide-based.

Step 2: Complete hydrolysis of 1.0 × 10⁻⁴ mol of C liberates 1.0 × 10⁻⁴ mol of a fragment that is stable to hot phosphoric acid, reduces Tollens’ reagent and Benedict’s solution → the fragment is a reducing sugar. In NADP⁺ the 2′-phosphate on the adenosine ribose is hydrolytically stable; after cleavage of the pyrophosphate and nicotinamide–ribose bonds this 2′-phospho-ribose (or ribose-5-phosphate) is released and behaves as a reducing sugar. Thus the extra phosphate of NADP⁺ is responsible for the observed reducing sugar.

Step 3: Exhaustive enzymatic de-phosphorylation (removes all phosphates) followed by acid hydrolysis of 1.0 × 10⁻⁴ mol C gives 2.0 × 10⁻⁴ mol NH₃ and 1.0 × 10⁻⁴ mol of a 6-carbon aldaric acid. Two mol NH₃ per mol C indicates two amino groups in the original molecule; these are the amide NH₂ of nicotinamide and the amino group of the adenine ring (both released on strong acid hydrolysis). The 6-carbon aldaric acid arises from the glucose/ribose skeleton. The acid is optically active and its meso-form is inactive → it must be an aldaric acid with a C2-axis, e.g. D-glucaric acid (obtained by nitric acid oxidation of D-glucose). L-idaric and L-gularic acids are not consistent with the symmetry argument.

Step 4: Combining the 260 nm absorbance, the extra phosphate (giving a reducing sugar on hydrolysis), and the formation of D-glucaric acid, the co-enzyme must be NADP⁺.

Conclusion: C is NADP⁺ and the 6-carbon aldaric acid is D-glucaric acid.
Question 5
4 / -1

The concentration of a reactant decreases from 0.2 M to 0.1 M in 10 minutes. The rate of the reaction is

A
0.01 M

B
10⁻²

C
0.01 mol dm⁻³ min⁻¹

D
1 mol dm⁻³,min⁻¹

Solution

Step-by-step solution

The rate of a reaction (average rate over the given interval) is defined as the change in concentration of a reactant per unit time, with a negative sign to make the rate positive:
Question 6
4 / -1

An aqueous solution of titanium chloride, when subjected to magnetic measurement, measured zero magnetic moment. Assuming the octahedral complex in aqueous solution, the formulae of the complex is:

A

B

C

D

Solution

Step 1: Zero magnetic moment ⇒ the complex is diamagnetic, i.e. contains no unpaired electrons.

Step 2: Titanium in an octahedral aqua complex must therefore adopt an oxidation state that gives a d⁰ configuration (the only Ti-based configuration with 0 unpaired electrons).

Step 3: Among the choices only [Ti(H₂O)₆]Cl₄ corresponds to Ti in the +4 oxidation state (Ti⁴⁺: 3d⁰), which is diamagnetic.

Step 4: All other options contain Ti(III) (3d¹) or Ti(II) (3d²), each of which would give a non-zero magnetic moment.

Hence the correct formula is [Ti(H₂O)₆]Cl₄.
Question 7
4 / -1

The pair in which both species have same magnetic moment (spin only value) is

A

B

C

D

Solution
Question 8
4 / -1

Equal masses of a solute are dissolved in equal amount of two solvents A and B, respective molecular masses being M_A and M_B . The relative lowering of vapour pressure of solution in solvent A is twice that of the solution in solvent B. If the solutions are dilute, M_A and M_B are related as

A
M_A= M_B

B
2M_A= M_B

C
M_A= 2M_B

D
M_A= 4M_B

Solution
Question 9
4 / -1

Which of the following is the sweetest sugar

A
Glucose

B
Fructose

C
Lactose

D
Sucrose

Solution

Given: The chemical formula of fructose is C₆H₁₂O₆.

It is the sweetest sugar.

It is a naturally occurring sugar.

It is a monosaccharide, meaning the simplest form of sugar, which means it cannot be broken down into more basic sugar units.
Question 10
4 / -1

Principal quantum number of an atom represents

A
Size of the orbital

B
Spin angular momentum

C
Orbital angular momentum

D
Space orientation of the orbital

Solution

The principal quantum number, denoted by n, is one of the four quantum numbers used to describe the state of an electron in an atom.

It primarily determines the electron's energy level and the average distance of the electron from the nucleus, which corresponds to the size of the orbital.

A larger value of n indicates a larger orbital size and higher energy.

Other quantum numbers:

Spin angular momentum is represented by the spin quantum number ( m_s).

Orbital angular momentum is related to the angular momentum quantum number ( l ).

Space orientation of the orbital is determined by the magnetic quantum number (m_l).