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NEET Chemistry MCQ
Quiz 3
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Q.1
The total number of electrons in 3 g of KH to that in 1.8 g of H2O
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a) Double
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b) Same
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c) Triple
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d) One fourth
Explanation
Molecular mass of KH=29+1=30g 3g of = 0.1 mole One molecule of KH have 19+1= 20 electrons ∴ 0.1 mole of KH have = 0.1×20 =2 mole of electrons Molecular mass of H2O= 2(1)+16 =18g 1.8g =0.1 mole One molecule of H2O have 2(1)+8 = 10 electron 0.1 mole of H2O have = 0.1×10 = 1 mole electrons The total number of electrons in 3 g of KH to that in 1.8 g of H2O is double Answer : (a)
Q.2
Change in volume when 100 mL PH3 decomposed to solid phosphorus and H2 gas.
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a) Increase in 50 mL
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b) Decrease in 50 mL
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c) Increase in 150 mL
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d) Decrease in 200 mL
Explanation
4PH3 → P4 + 6H2( as Phosphorus is Tetra atomic molecules) 4 mL of PH3 gives 6mL of H2 ∴ 100 mL of PH2 gives 150mL of H2 Increase in volume = 150-100 =50 mL Answer : (a)
Q.3
An element, X has the following isotopic composition 200X : 90% ; 199X : 8%; 202X : 2.0%. The weighted average atomic mass of the naturally occurring element X is closest to
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a) 201 amu
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b) 202 amu
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c) 199 amu
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d) 200 amu
Explanation
Using formula for average atomic mass Average mass=(200×90+199×8+202×2)/100 On calculating we get Average atomic mass = 199.08 ≈ 199amu Answer : (c)
Q.4
When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g), each at STP, the moles of HCl(g) formed is equal to [AIPMT-2014]
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a) 1 mol of HCl(g)
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b) 2 mol of HCl(g)
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c) 0.5 mol of HCl(g)
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d) 1.5 mol of HCl(g)
Explanation
Reaction is H2+ Cl2 → 2HCl One of each H2 and C2 produce 2 moles of HCl 22.4 L of H2= 1 mole 11.2 Lof Cl2 = 0.5 mole As Cl2 is only 0.5 mole ∴HCl produced will be 1 mole and Hydrogen of 0.5 mole will remain unreacted Answer : (a)
Q.5
ssertion: 50 ml, decimolar H2SO4 when mixed with 50 ml, decimolar NaOH, then normality of resultant solution is 0.05 N. Reason: Here, NV = | N1V1 – N2V2|.
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a) Both Assertion and Reasoning are true and Reasoning is correct explanation of Assertion
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b) Both are correct but Reason is not explanation of Assertion.
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c) Assertion is correct but Reason is wrong.
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d) Both Assertion and Reason are incorrect.
Explanation
Answer : (a)
Q.6
Assertion : 1 gram of salt in 1 m3 of solution has concentration of 1 ppm. Reason: ppm is defined as number of parts by mass of solute per million parts of solution.
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a) Both Assertion and Reasoning are true and Reasoning is correct explanation of Assertion
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b) Both are correct but Reason is not explanation of Assertion.
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c) Assertion is correct but Reason is wrong.
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d) Both Assertion and Reason are incorrect.
Explanation
Answer : (a)
Q.7
When 100 ml of (M/10) H2SO4 is mixed with 500 ml of (M/10) NaOH then nature of resulting solution and normality ofexcess of reactant left is
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a) Acidic, N /5
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b) Basic,N/5
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c) Basic, N /20
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d) Acidic,M/10
Explanation
Reaction is H2SO4 + 2NaOH → Na2SO4+ 2H2O One mole of H2SO4 reacts with 2 moles of NaOH Since both the solution have same concentration 100 ml of H2SO4 will neutralize 200 mLof NaOH Thus 300 mL of NaOH of 0.1M will remain reacted =0.03mole of NaOH But after adding both the reactant volume of solution= 100+ 500 = 600 mL And mixture have 0.03 mole of NaOH ∴molarity of NaOH = 600mL =0.03 mole 1000 mL = N/20 ∴ mixture will be basic, N/20 Answer: (b)
Q.8
The number of atoms in 0.1 mol of a triatomic gas is (NA=6.02× 1023mol–1) [AIPMT (Prelims)-2010]
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a) 6.026 × 1022
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b) 1.806 × 1023
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c) 3.600 × 1023
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d) 1.800 × 1022
Explanation
Number of atoms in one mole of triatomic gas= 3 . Number of atoms = 3 × 0.1 × 6.022 × 1023 = 1.806 × 1023 Answer (b)
Q.9
Ammonia gas is passed into water, yielding a solution of density 0.93 g/cm3 and containing 18.6% NH3 byweight. The mass of NH3 per cc of the solution is
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a) 0.17 g/cm³
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b) 0.34 g/cm³
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c) 0.51 g/cm³
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d) 0.68 g/cm³
Explanation
Density is0.93g/cm3 or 0.93g/mL Thus weight of 1mL NH3 solution is 0.93g Since NH3solution if of 18.6% by weight ∴ 18.6%of 0.93 g = 0.1728 g ≈ 0.17g/cm3 Answer(a)
Q.10
What volume of oxygen gas (O2) measured at 0°C and 1 atm, is needed to burn completely 1 L of methane gas (CH4) measured under the same conditions?
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a) 1 L
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b) 2 L
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c) 3 L
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d) 4 L
Explanation
One mole of gasses occupy equal volume irrespective of there atomic masses CH4 + 2O2 → CO2 + 2H2O Or one mole of CH4 requires two moles of oxygen Thus 1L of CH4 requires 2 L of oxygen Answer : (b)
Q.11
Mole fraction of solvent in aqueous solution of NaOH having molality of 3 is
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a) 0.3
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b) 0.05
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c) 0.7
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d) 0.95
Explanation
Molality = moles per Kg of solvent Given 3 moles of NaOH and solvent is water (aqueous)= 1Kg= 55.5 Thus mole faction x x=55.5/(3+55.5)=0.9487 Answer: (d)
Q.12
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced inthis reaction will be [AIPMT (Prelims)-2009]
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a) 3 mol
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b) 4 mol
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c) 1 mol
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d) 2 mol
Explanation
Reaction 2H2 (g) + O2 (g) → 2H2 O(l) Two moles of H2 and one mole of O2 produce two moles of water 10g of hydrogen=5 moles 64g of oxygen = 2 moles ∴oxygen is limiting reagent , Now 2 moles will produce 4 moles of water Answer (b)
Q.13
What volume of oxygen gas (O2) measured at 0°C and 1 atm, is needed to burn completely 1 L of propanegas (C3H8) measured under the same conditions? [AIPMT ((Prelims)-2008]
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a) 10 L
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b) 7 L
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c) 6 L
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d) 5 L
Explanation
Reaction C3H8 + 5O2 → 3CO2 + 4H2O Thus 1 L requires 5Lof O2 Answer: (d)
Q.14
The total number of electrons in 2.0 g of D2O to that in 1.8 g of H2O
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a) Double
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b) Same
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c) Triple
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d) One fourth
Explanation
Both atoms D2O and H2O have same number of electron Molecular weight of D2O = 20g Molecular weight of H2O =18 g Thus total number of electrons in 2.0 g of D2O to that in 1.8 g of H2O are same as number of moles are same = 0.1 mole Each H2O have 10 electrons Thus number of electron =1mole Answer: (b)
Q.15
Assertion : 50 ml, decinormalHCl when mixed with 50 ml, decinormal H2SO4, then normality of H+ ion in resultantsolution is 0.1 N. Reason : Here, MV = M1V1+ M2V2
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a) Both Assertion and Reasoning are true and Reasoning is correct explanation of Assertion
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b) Both are correct but Reason is not explanation of Asserion.
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c) Asserion is correct but Reason is wrong.
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d) Both Asserion and Reason are incorrect.
Explanation
Answer : (a)
Q.16
How many litre of oxygen at STP is required to burn 60 g C2H6?
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a) 22.4 L
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b) 11.2 L
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c) 22.4 × 7 L
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d) 8.5 L
Explanation
Molar mass of C2H6 = 24+6 = 30g 60g of C2H6= 2 moles Burning of C2H6 equation is 2C2H6 + 7O2 → 4CO2 + 6H2O Thus 2 moles of C2H6 required 7 moles of O2 Now volume of 1 mole of gas at STP = 22.4 L Volume of 7 moles of O2= 22.4 × 7 L Answer: (c)
Q.17
If Avogadro number NA, is changed from 6.022 × 1023mol-1 to 6.022 × 1020mol-1, this would change[Re-AIPMT-2015]
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a)The ratio of chemical species to each other in a balanced equation
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b) The ratio of elements to each other in a compound
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c) The definition of mass in units of grams
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d) The mass of one mole of carbon
Explanation
Atomic mass unit is one twelfth of mass of C-12. A mass of one mole C = Avogadro number × mass of one atom of C Answer: (d)
Q.18
How many g of dibasic acid (mol. weight 200) should be present in 100 ml. of the aqueous solution to give strength of 0.1 N?
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a) 10 g
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b) 2 g
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c) 1 g
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d) 20 g
Explanation
Dibasic acid liberate two H+ ions which is required for calculation of Normality For given solute molecular wt = 200 ∴ Equivalent weight = 200/2= 100 Normality is given by formula ∴ w = 1 g Answer: (c)
Q.19
Percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (at. wt. = 78.4) then minimum molecular weight of peroxidase anhydrous enzyme is
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a) 1.568 × 104
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b) 1.568 × 103
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c) 15.68
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d) 2.136 × 104
Explanation
Minimum one atom Of Sc must be present in given compound Now 0.5%= 78.4 ∴ 100% = 15680 = 1.568×104 Answer: (a)
Q.20
An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H,9.67%. The empirical formula of the compound would be [AIPMT (Prelims)-2008]
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a) CH4O
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b) CH3O
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c) CH2O
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d) CHO
Explanation
% oxygen = 100 - 38.71 – 9.67 = 61.62% C = 38.71 ∴ Number of atoms 38.71/12 = 3.22 H = 9.67 ∴ Number of atoms 9.67/1 = 9.67 Thus Hydrogen atoms are three times the carbon Option b is correct Answer: (b)
0 h : 0 m : 1 s
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