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Test: Chemistry - 7 - MCAT MCQ


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15 Questions MCQ Test - Test: Chemistry - 7

Test: Chemistry - 7 for MCAT 2024 is part of MCAT preparation. The Test: Chemistry - 7 questions and answers have been prepared according to the MCAT exam syllabus.The Test: Chemistry - 7 MCQs are made for MCAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Chemistry - 7 below.
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Test: Chemistry - 7 - Question 1

How many significant figures are there in the following figures?
i. 6*104
ii. 0.008320
iii. 4.05*10-2
iv. 100.0

Detailed Solution for Test: Chemistry - 7 - Question 1

6*104 = 6000 has only one significant figure. Leading zeros are not significant, for 0.008320 it is 4. Zeros appearing anywhere between two non-zero digits are significant figures, for 4.05*10-2 = 0.00405, it is 3. Trailing zeros in a number containing a decimal point are significant, for 100.0 it is 4.

Test: Chemistry - 7 - Question 2

The conversion of liquid to solid is known as

Detailed Solution for Test: Chemistry - 7 - Question 2

Freezing – liquid to solid, melting – solid to liquid, sublimation – solid to vapour, condensation – gas to liquid, deposition – gas to solid.

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Test: Chemistry - 7 - Question 3

In a polyatomic species, the sum of oxidation numbers of the element in the ion _________ thecharge on that species.

Detailed Solution for Test: Chemistry - 7 - Question 3

The sum of oxidation numbers in polyatomic ion or species is equal to the charge of the ion. For example, the sum of the oxidation number for SO42- is -2.

Test: Chemistry - 7 - Question 4

The hybridisation in NH4is

Detailed Solution for Test: Chemistry - 7 - Question 4

Number of valence electrons in N is 5 and in H it is 4.
So total number of valence electrons = 5 + 4 = 9; Charge = +1.
Therefore, total electrons in NH4+ = 9 - 1 = 8
When the total number of electrons is less than 8, divide by 2. If it lies between 9 and 56, divide it by 8.
8/2 = 4; X = 4
Therefore, hybridisation in NH4+ is sp3.

Test: Chemistry - 7 - Question 5

Which of the following solvents is suitable for SN2 reactions?

Detailed Solution for Test: Chemistry - 7 - Question 5

Aprotic solvents do not solvate the anions effectively and it is used for SN2 reactions. Acetonitrile is the only aprotic solvent whereas others are polar protic solvents.

Test: Chemistry - 7 - Question 6

The base peak in a mass spectrum is

Detailed Solution for Test: Chemistry - 7 - Question 6

The most intense peak is called as base peak. It usually corresponds to the molecular ion only, if the spectra are recorded at low ionization energy.

Test: Chemistry - 7 - Question 7

Which of the following shows the increasing order of solubility?

Detailed Solution for Test: Chemistry - 7 - Question 7

KCl is highly soluble because its solubility is greater than 0.1M. AgCl is sparingly soluble because its solubility is less than 0.01 M. PbS is least sparingly soluble becauseits solubility is very much less than 0.01 M.

Test: Chemistry - 7 - Question 8

An ideal gas can be defined thermodynamically, when,
I. PV = constant
II. (∂U/∂V)p = 0
III. (∂U/∂V)T = 0

Detailed Solution for Test: Chemistry - 7 - Question 8

For an ideal gas, PV = constant, at constant temperature. The internal energy of a given quantity of an ideal gas at a constant temperature is independent of its volume, thus (∂U/∂V)T = 0.

Test: Chemistry - 7 - Question 9

One mole of any gas at STP occupies

Detailed Solution for Test: Chemistry - 7 - Question 9

By applying Ideal gas equation, V = nRT/P
At STP, P = 1 atm, n = 1mol, R = 0.082 L atm K-1 mol-1, T = 273K
V = (1*0.082*273)/273 = 22.38 L = 22.4 L

Test: Chemistry - 7 - Question 10

Identify the unit of concentration of the solution (NA)/(Kg of solvent).

Detailed Solution for Test: Chemistry - 7 - Question 10

Molarity (MA) = nA/ volume in litres.
Normality = Gram equivalent of A/Volume in litres of solution.
Mole fraction (χi) = ni/ (n1+n2+n3....).
Parts per million (ppm) = (Mass of A/Total mass) x 106

Test: Chemistry - 7 - Question 11

In which of the following processes, is the process always non-feasible?

Detailed Solution for Test: Chemistry - 7 - Question 11

For a non-spontaneous or non-feasible process, ΔH > 0 and ΔS < 0. For a spontaneous or irreversible reaction, ΔH < 0 and ΔS > 0. For an equilibrium or reversible process, ΔH = 0 and ΔS = 0.

Test: Chemistry - 7 - Question 12

Slater's rule is used to calculate the value of

Detailed Solution for Test: Chemistry - 7 - Question 12

The value of screening constant (S) and effective nuclear charge (Z*) can be calculated using Slater's rule. Effective charge (Z*) = Z – S (where Z- atomic number and S-screening constant).

Test: Chemistry - 7 - Question 13

Identify the glass equipment with ground-glass joints

Detailed Solution for Test: Chemistry - 7 - Question 13

Glass equipments are divided into two; with ground-glass joints and without ground-glass joints. Separating funnel is the only glass equipment with ground-glass joints.

Test: Chemistry - 7 - Question 14

Which of the following is the weakest base?

Detailed Solution for Test: Chemistry - 7 - Question 14

The electronegativity and atomic size of iodine is larger so there is a weaker bond between hydrogen and iodine that makes the electron cloud much lesser than H-F bond. So, H-I is the weakest base; in other words it is the strongest acid.

Test: Chemistry - 7 - Question 15

Calculate the cell potential at 250C for the following cell reaction using Nernst equation. Eoox = -3.402 V, E0red = 0.7996 V
Cu|Cu2+(0.024 M)||Ag+(0.0048 M)|Ag

Detailed Solution for Test: Chemistry - 7 - Question 15

Oxidation: Cu→Cu2+ + 2 e- Eoox = -(0.340 V) Reduction: Ag+ + e- → Ag Eored = 0.799 V Overall cellreaction is Cu(s) + 2 Ag+(aq)→ Cu2+ (aq) + 2 Ag(s) E0cell = E0red + E0ox = 0.799 V + (-0.340 V) = 0.459V Nernst equation, Ecell = E0cell – (0.0256/n) (Inox/Inred) = 0.459 –(0.0256/2) * In [0.024 / (0.0048)2] = 0.459 – 0.0128 * In (1043) = 0.459 – 0.0128 * 6.95 Ecell = 0.370 V

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