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NEET Chemistry MCQ
Quiz 4
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Q.1
In Haber process 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture under the aforesaid conditionin the end?
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a) 20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen
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b) 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
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c) 20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen
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d) 20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
Explanation
Reaction N2 + 3H2 → 2NH3 Hydrogen is 30 L thus requires 10 Nitrogen Hydrogen is limiting reagent Now only 50% of Hydrogen is used = 15 and Balance hydrogen =15L Therefore 5 L of Nitrogen used and 30-5 = 25 Balanced NH3 Produced = 10 L Answer: (b)
Q.2
2.5 litre of 1 M NaOH solution mixed with another 3 litre of 0.5 M NaOH solution. Then find out molarity ofresultant solution.
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a) 0.80 M
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b) 1.0 M
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c) 0.73 M
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d) 0.50 M
Explanation
Mixing solutions are same M1V1 + M2V2 = M(V1+V2) 1×2.5+0.5×3 = M(2.5+3) M= 0.7272 =0.73M Answer: (c)
Q.3
How many grams of CH3OH should be added to water to prepare 150 ml solution of 2 M CH3OH?
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a) 9.6 × 103
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b) 2.4 Times; 103
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c) 9.6
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d) 2.4
Explanation
Molecular weight of CH3OH = 12+4+16 = 32g M=(Wt of solvant×1000)/(Molar wt×volume of olution in mL) 2=(w×1000)/(32×150) W= 9.6 g Answer: (c)
Q.4
1 mol of KClO3 is thermally decomposed and excess of aluminium is burnt in the gaseous product. How many moles of Al2O3 are formed?
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a) 1
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b) 2
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c) 1.5
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d) 3
Explanation
Thermal decomposition of KClO3 2KClO3 → 2KCl +3O2 Thus Two moles of KClO3 produces Three moles of oxygen 1 mole of KClO3 will produce 1.5moleof Oxygen Now when Al burn 4Al +3O2 → 2Al2O3 Al is in excess thus oxygen is limiting reagent Now 3 moles of Oxygen produce 2 moles of Al2O3 × 1.5 moles of Oxygen produce 1 mole of Al2O3 Answer : (a)
Q.5
A mixture of gases contains H2 and O2 gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the two gases in the mixture? [AIPMT-2015]
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a) 2 : 1
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b) 1 : 4
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c) 4 : 1
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d) 16 : 1
Explanation
If 5 g is mixture H2 = 1g = 0.5 mole( as molecular weight of H2= 2g) O2 = 4g = 0.125 moles( as molecular weight of O2 = 32g) Thus ration 0.5:0.125 =4:1 Answer: (c)
Q.6
Assertion: Number of ions in 9 gram ofNH4 is equal to Avogadro's number (NA). Reason: Number of ions is equal to number of atoms.
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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 : (d)
Q.7
1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reactant is left in excess and how much? (At. wt. Mg = 24; O = 16) [AIPMT-2014]
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a) Mg, 0.16 g
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b) O2, 0.16 g
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c) Mg, 0.44 g
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d) O2, 0.28 g
Explanation
Reaction is 2Mg + O2 → 2MgO Two moles of Mg reacts with one mole of oxygen to produce 2moles of MgO 1.0g of Mg = 1/24 moles = 4/96 0.56 g of O2 = 0.56/32 moles = 1.68/96 For 1.68/96 moles of oxygen Mg requires = 2(1.68/96) Mg moles taken is 4/96 which is more than complete reaction of O2 Excess moles Mg=4/96 -(2×1.68)/96=0.64/96 Excess weight of Mg= Excess mole × molecular wt Excess weight =0.64/96×24=0.16g Answer : (a)
Q.8
The total number of electrons in 4 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=39+1=40g 4g 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 4 g of KH to that in 1.8 g of H2O is double Answer : (a)
Q.9
20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesiumoxide. What will be the percentage purity of magnesium carbonate in the sample? (At. wt. : Mg = 24) [Re-AIPMT-2015]
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a) 60
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b) 84
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c) 75
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d) 96
Explanation
MgCO3 → MgO + CO2 One mole of MgCO3 gives one mole each of MgO and CO2 Molecular weight of MgCO3 = 24+12+3(16) = 84g Molecular weight of MgO = 24+16 = 40g 40g of MgO = 84 MgCO3 ∴ 8 g of MgO = (84/5) g MgCO3 = 16.8 g But sample is 20g 20g sample = 16.8 MgCO3 100g same = 16.8 × 5 = 84 MgCO3 Answer : (b)
Q.10
Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 gmL–Volume of acid required to make one litre of 0.2 M H2SO4 solution is
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a) 16.65 mL
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b) 22.20 mL
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c) 5.55 mL
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d) 11.11 mL
Explanation
To make 0.1 M of H2SO4 Molecular weight of H2SO4 is 98 g ∴ 19.6 g of sulphuric acid require to make 1L of 0.2M solution As purity is 98% thus acid required will be =20 g Now density is 1.8gmL-1 therefore weight of 1000mL acid = 1800 g ∴ 20g = 11.11 mL Alternate method you can use formula for M if density and %w/w given M=(%w/w × density ×10)/(molecular mass) Then use M1V1 = M2V2 Answer (d)
Q.11
Volume occupied by one molecule of water (density = 1 g cm–3) is [AIPMT (Prelims)-2008] Solution
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a) 5.5 × 10⁻²³ cm³
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b) 9.0 × 10⁻²³ cm³
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c) 6.023 × 10⁻²³ cm³
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d) 3.0 × 10⁻²³ cm³
Explanation
As density =1g/cc 55.5 moles of water occupy 1000 mL volume 55.5×6.023×1023 molecules = 1000 mL Volume 1 molecule = 0.00299×10⁻²⁰ L Volume ∴ Volume of 1 molecule = 3×10⁻²³ cm3 Answer : (d)
Q.12
Rearrange the following ( I to IV ) in the order of increasing masses and choose the correct answer from (a), (b), (c) and (d) ( Atomic mass: N=14, O=16, Cu=63)I. 1 molecule of oxygenII. 1 atom of nitrogenIII. 1×10⁻¹⁰ g molecular weight of oxygenIV. 1×10⁻¹⁰ g atomic weight of copper[ IIT 1993]
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a) II < I < III < IV
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b) IV < III < II < I
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c) II < III < I < IV
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d) III < IV < I < II
Explanation
I. 1 molecule of oxygen=32 / NA=5× 10⁻²³ II. 1 atom of N=14/NA=2.3×10⁻²³III. 10-10g mol wt. of oxygen=10-10 ×32=3.2×10⁻⁹g IV. 10-10g atom of copper=10-10×63=6.3×10⁻⁹Answer: (a)
Q.13
Rearrange the following in increasing mass I ) One molecule of Oxygen II) One atom of Nitrogen III) 10-10 g molecular weight of oxygen IV) 10-10 g atomic weight of copper
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a) II < I < III < IV
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b) IV < III < II < I
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c) II < III < I < IV
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d) III < IV < I < II
Explanation
Mass of One molecule of Oxygen = 32/NA Mass of One atom of Nitrogen = 14/NA Mass of 10-10 g molecular weight of oxygen = 32× 10⁻¹⁰ g Mass of 10-10 g atomic weight of copper = 64× 10⁻¹⁰ g Answer : (a)
Q.14
Specific volume of cylindrical virus particle is 6.02 × 10⁻² cc/gm whose radius and length are 7 Å and 10 Å respectively. If NA= 6.02 ×1023, find molecular weight of virus.
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a) 15.4 kg/mol
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b) 1.54 × 104 kg/mol
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c) 3.08 × 104 kg/mol
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d) 3.08 × 103 kg/mol
Explanation
1 Å = 10-10 m Volume of virus = πr2h = π ×(7×10⁻¹⁰)2× 10×-10 cc = 1.54×10⁻²⁷ m3 6.02× 10⁻² cc = 1 g Or 6.02 × 10⁻⁸ m3 = 10-3 kg ∴ weight of 1.54 × 10⁻²⁷ m- = w W is the weight of one virus ∴molecular weight M of virus = w ×NA Answer : (a)
Q.15
Number of Fe atoms in 100 g Haemoglobin if it contains 0.33% Fe. (Atomic mass of Fe = 56)
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a) 0.035 ×1023
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b) 35
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c) 3.5 ×1023
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d) 7 × ×108
Explanation
Given 0.33% Fe of 100gHaemoglobin = 0.33g Molecular mass of Fe = 56 i.e. 56g = 1 mole ∴ 0.33 g = 0.33/56 moles = 5.9 ×10⁻³ ∴ number molecules of Fe = 5.9 ×10⁻³× 6.023×1023 = 0.03553 ×1023 Atoms Answer: (a)
Q.16
Assertion : Total charge on NA ions of CO32- is 1.93 × 105 coulomb. Reason: Charge on one electron in 96500 coulomb
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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
As one CO3-2 ion have two unit charge ∴ 1 mole of CO3-2 ion havee charge = 2 × 96500 C = 1.93 × 105 C Charge on one mole electron = 1.602 × 10⁻¹⁹× 6.022 × 1023 =96478 = 96500 C Answer : (c)
Q.17
Volume occupied by one molecule of water (density = 1 g cm–3) is …
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a) 5.5 × 10⁻²³ cm³
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b) 9.0 × 10⁻²³ cm³
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c) 6.023 × 10⁻²³ cm³
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d) 3.0 × 10⁻²³ cm³
Explanation
Mass of one mole of water or molecular mass =18g As volume = mass/density ∴ Volume of 1 mole of water = 18 ml One mole contains 6.023× 10⁻²³ number of molecules ∴ Volume of one molecule of water = 18/(6.023×1023 )=3×10⁻²³ ml As 1ml = 1cm3 ∴ Volume occupy by one molecule of water = 3.0 × 10⁻²³ cm3 Answer : (d)
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