45 Questions MCQ Test NEET Mock Test Series - Test: Chemistry - 1
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Two oxides of metal contain 27.6℅ and 30℅ oxygen respectively. If formula of first oxide is M3O4, then formula of second oxide is:
Detailed Solution for Test: Chemistry - 1 - Question 1
Given that, formula of first oxide = M3O4
Let mass of the metal = x
% of metal in M3O4 = (3x / 3x + 64) * 100
but as given % age = (100 - 27.6) = 72.4%
so, (3x / 3x + 64) * 100 = 72.4
or x = 56.
in 2nd oxide,
oxygen = 30%....so metal = 70%
so, the ratio is
M : O
70 / 56 : 30 / 16
1.25 : 1.875
2 : 3
so, 2nd oxide is M2O3
The crystalline salt, Na2SO4xH2O on heating loses 55.9% of its weight. The formula of the crystalline salt is
Detailed Solution for Test: Chemistry - 1 - Question 3
Molar mass of salt = (2 × 23) + 32 + (16 × 4) + x × 18
= 142 + 18x
On heating it will lose water and become anhydrous. 55.9% mass is the mass of water in Na2SO4.xH2O.
Mass by mass% = (mass / molar mass of compound) × 100
mass of water = 18x
55.9 = (18x / 142 + 18x) × 100
55.9(142 + 18x) = 1800x
7937.8 + 1006.2x = 1800x
x = 9.99
x = 10 (approx.)
So, the molecular formula of compound is Na2SO4.10H2O.
Air contains nearly 20% oxygen by volume. The volume of air needed for complete combustion of 100 cc of acetylene will be
Detailed Solution for Test: Chemistry - 1 - Question 5
2 C2H2 + 5O2 → 4CO2 + 2 H2O
In this reaction, we can see that, 2 moles of C2H2reacts with 5 moles of O2
Or, 1 mole of C2H2 reacts with 5 / 2 moles of O2
Under standard conditions, we can say that:
22.44 L of C2H2 reacts with (5 / 2) x 22.4 L moles of O2
So, 100 cc of C2H2 will reacts with (5 / 2) x 100 cc = 250 cc of O2 .
But, since air contains 20 % of oxygen by volume, the amount of air needed to react with 100 cc of C2H2 will be = 250 x (100 / 20) = 1250 cc of air.
A solution of 10 ml(M/10) FeSO4 was titrated with KMnO4 solution in acidic medium. The amount of KMnO4 used will be
Detailed Solution for Test: Chemistry - 1 - Question 6
The equation of the Mn2+ in acidic medium will be.
MnO4- + 8H++ 5 Fe2+ → Mn2++ 5Fe3+ +4H2O.
We can balance the reaction by using redox balancing of acid and base.
So, we know from the equation that 5 moles of iron react to form 1 mole of Mn+2.
Thus, we can say that (10*0.1) moles of iron will react with the (10*0.1/5) moles of Manganese ions, which will be 0.2 thus, 10ml volume of KMnO4 and 0.02M moles of KMnO4. We will get the same result if we use M1V1 = M2V2 formulae.
Detailed Solution for Test: Chemistry - 1 - Question 8
Dalton's law of multiple proportions is part of the basis for modern atomic theory, along with Joseph Proust's law of definite composition (which states that compounds are formed by defined mass ratios of reacting elements) and the law of conservation of mass that was proposed by Antoine Lavoisier.
Detailed Solution for Test: Chemistry - 1 - Question 9
The discovery of protons can be attributed to Rutherford. In 1886 Goldstein discovered existence of positively charged rays in the discharge tube by using perforated cathode. These rays were named as anode rays or cannal rays.
The number of unpaired electrons in Cu+ (Z= 29) is
Detailed Solution for Test: Chemistry - 1 - Question 14
Answer : The electronic configuration of Cu is,
The reason for abnormal electronic configuration of Cu is because of the half-filled and fully-filled configuration are the more stable configuration. In this configuration, the d-subshell is fully-filled and s-subshell is half-filled which makes the Cu more stable.
Now, for getting a positive charge, it has to lose 1 electron which will be removed from the outermost 4s orbital and hence all the remaining electrons will be paired leading to 0 unpaired electrons.
Consider the following statements. The d-orbitals have
I. four lobes and two nodes
II. four lobes and one node
III. same sign in the opposite lobes
IV. opposite sign in the opposite lobes
Which of the above statements are correct ?
Detailed Solution for Test: Chemistry - 1 - Question 16
Total no of angular nodes = l (azimuthal quantum no), so l=2 in d orbital, therefore number of nodes is 2 and it is a fact that opposite lobes have same signs and adjacent ones have opposite signs. there are 4 lobes in d orbital.
Which of the following substances has the lowest boiling point?
Detailed Solution for Test: Chemistry - 1 - Question 17
Boiling Point is directly proportional to VanderWaals forces of attraction. Down the group, this forces increases. so force of attraction in CHF3 is weaker than CHL3, that's why boiling point of CHF3 is lowest.
Out of the two compounds shown below, the vapour pressure of B at a particular temperature is expected to be
Detailed Solution for Test: Chemistry - 1 - Question 19
Due to intramolecular H-bonding in o-nitrophenol it exists as a monomer while due to intermolecular H-bonding in p-nitrophenol it exists as an associate molecule. As a result, vapour pressure of a o-nitrophenol is higher than that of a p-nitrophenol.
In which of the following molecules, the central atom does not have sp3 hybridisation?
Detailed Solution for Test: Chemistry - 1 - Question 20
When the number of hybrid orbitals, H is 4, the hybridization is sp3 hybridization
H = 1 / 2[V + M - C + A]
where, V = number of monovalent atoms
C = total positive charge
A = negative charge
a) For CH4
H = 1 / 2[4 + 4 - 0 + 0] = 4
b) For SF4,
H = 1 / 2 [6 + 4 - 0 + 0] = 5
c) For BF-4,
H = 1 / 2 [3 + 4 - 0 + 1] = 4, thus sp3
d) For NH+4
H = 1 / 2[5 + 4 - 1 + 0] = 4 thus sp3
Thus, only in SF4 the central atom does not have sp3 hybridisation.
The correct oreder of increasing bond angles in the following species
Detailed Solution for Test: Chemistry - 1 - Question 21
According to VESPR theory repulsion order
lp - lp > lp - bp > bp - bp
As the number of lone pairs of electrons increases, bond angle decreases due to repulsion between lp-lp, Moreover, as the electronegativity of central atom decreases, bond angle decreases.
The correct order of increasing bond angle is Cl2O < ClO2- < ClO2
In ClO2- there are 2 Ione pairs of electrons present on the central chlorine atom. Therefore the bond angle in ClO2- is less than 118° which is the bond angle in ClO2 which has less number of electrons on chlorine.
Detailed Solution for Test: Chemistry - 1 - Question 24
Li+ have larger size in aq.solution due to the presence of the water molecules present inside the solution surronds easily and become the solution to be more hydrated comparing to the corresponding element present in the aqeous solution.
Detailed Solution for Test: Chemistry - 1 - Question 26
On moving down the group, there is an increase in the size of atoms. As a result, the tendency to accept an additional electron decreases and hence, the electron gain enthalpy decreases from Ne to Rn. However, He has small size and much higher tendency to accept an additional electron. Hence, its electron gain enthalpy is least positive of all the noble gases.
Thus, it is concluded that neon has the highest positive electrons gain enthalpy.
The electron gain enthalpies of B,C,N and O are in the order
Detailed Solution for Test: Chemistry - 1 - Question 27
Actually electron gain enthalphy and ionisation energy both are dependent on electronic configuration.
Electronegativity means tendency to attract electron it doesnot matter where it will keep the electron
Electron gain entalphy basically refer to attract e it must have space to keep electron in sub shell like spdf should be present
order of electronegativity is
F > O > N > Cl > Br > I > S > C > P
Detailed Solution for Test: Chemistry - 1 - Question 28
With respect to Chlorine hydrogen will be electropositive. Chlorine being a strong electronegative element tends to force hydrogen to loss electron and hydrogen acquire electropositive character so that chlorine can complete its octet .
Which of the following decomposes at highest temperature ?
Detailed Solution for Test: Chemistry - 1 - Question 33
BaCO3 decomposes at highest temp.
All the carbonates decompose on heating to give CO2 and metal oxide.
The stability of carbonate towards heat depends upon the stability of the resulting metal oxide. More is the stability of the resulting metal oxide lesser is the stability of the carbonate towards heat and vice versa.
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