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NEET Chemistry MCQ
Chemical Arithmetic And Volumetric Analysis Mcq Neet Chemistry
Quiz 1
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Q.1
The volume occupied by 4.4g of CO2 as STP is ... [ AFMC 1997]
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a) 22.4L
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b) 2.24L
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c) 0.224L
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d) 0.1L
Explanation
molecular weight of CO2 is 44g 4.4g=0.1 mole1 mole at STP=22.4L0.1 mole at STP=2.24LAnswer: (b)
Q.2
In order to prepare one litre normal solution of KMnO4, how many grams of KMnO4 are required if the solution is to be used in acidic medium for oxidation? [ MP PET 2002]
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a) 158g
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b) 31.6g
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c) 62.0g
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d) 790g
Explanation
Answer: (b)
Q.3
Which one of the followings has maximum number of atoms ? [NEET 2018]
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a) 1 g of O2(g) [Atomic mass of O = 16]
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b) 1 g of Li(s) [Atomic mass of Li = 7]
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c) 1 g of Ag(s) [Atomic mass of Ag = 108]
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d) 1 g of Mg(s) [Atomic mass of Mg = 24]
Explanation
Answer: (b)
Q.4
100ml of PH3 when decompsed produces phosphorus and hydrogen. The change in volume is .... [ CPMT 1986]
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a) 50 ml increase
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b) 500 ml decrease
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c) 900 ml decrease
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d) nil
Explanation
2PH3 → 2P + 3H2 P is solid, And two moles of PH3 decomposes to produce 3 mole of hydrogen Thus 100 ml will produce 150ml of hydrogenIncrease is 50ml Answer: (a)
Q.5
Which of the following has smallest number of molecules? [ CPMT 1987]
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a) 0.1 mol of CO2 gas
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b) 11.2 L of CO2 gas at N.T.P.
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c) 22 g of CO2 gas
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d) 22.4×103 ml of CO2 gas
Explanation
11.2 L of CO2 gas at N.T.P.=0.5 mole 22 g of CO2 gas=0.5 mole22.4×103 ml of CO2 gas=1 moleAnswer: (a)
Q.6
At S.T.P. the density of CCl4 vapour in g/L will be nearest to .. [ CBSE PMT 1988]
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a) 6.84
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b) 3.42
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c) 10.26
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d) 4.57
Explanation
Molecular weight of CCl4=154 g at STP volume of one mole of gas=22.4 litdensity=mass/volume=154/22.4=6.875 g/LAnswer: (a)
Q.7
One litre hard water contains 12.00mg Mg2+ Milli-equivalents of washing soda required to remove its hardness is ... [ CBSE PMT 1988]
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a) 1
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b) 12.16
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c) 1×10-3
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d) 12.16×10-3
Explanation
Mg2+ + Na2CO3 → MgCO3 + 2Na+Thus 1 g eq of Mg2+ required 1 g eq of washing soda1g equivalent of Mg2+=12000 mgThus 12mg Mg2+ required=1g eq/1000=1 mili eq Na2CO3Answer: (a)
Q.8
Which property of an element is always a whole number? [ MP PMT 1986]
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a) Atomic weight
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b) Equivalent weight
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c) Atomic number
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d) Atomic volume
Explanation
Answer: (c)
Q.9
Two samples of lead oxide were separately reduced to metallic lead by heating in a current of hydrogen. The weight of lead from one oxide was half the weight of lead obtained from the other oxide. The data illustrates.. [ AMU 1983]
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a) law of reciprocal proportions
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b) law of constant proportions
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c) law of multiple proportions
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d) law of equivalent proportions
Explanation
Answer: (c)
Q.10
The percentage of copper and oxygen in samples of CuO obtained by different methods were found to be the same. This illustrates the law of ... [ AMU 1982]
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a) constant proportions
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b) conservation of mass
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c) multiple proportions
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d) reciprocal proportions
Explanation
Answer: (a)
Q.11
What quantity of lime stone ( CaCO3) on heating will give 56kg of CaO? [ DPMT 1981]
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a) 1000 kg
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b) 56 kg
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c) 44 kg
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d) 100 kg
Explanation
CaCO3 → CaO + CO2 Molecular weight of CaCO3=100 gmMolecular weight of CaO=56 Answer: (d)
Q.12
If two compounds have the same empirical formula but different molecular formulae, they must have .. [ MP PMT 1986]
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a) different percentage composition
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b) different molecular weights
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c) same viscosity
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d) same vapour density
Explanation
Answer: (b)
Q.13
1.520 g of the hydroxide of metal on ignition gave 0.995g of oxide. The equivalent weight of metal is ... [ DPMT 1984]
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a) 1.520
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b) 0.995
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c) 19.0
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d) 9.0
Explanation
on solving above equation we get E=9Answer: (d)
Q.14
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.15
The percentage of P2O5 in diammonium hydrogen phosphate, (NH4)2HPO4 is .... [ CPMT 1992]
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a) 23.48
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b) 46.96
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c) 53.78
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d) 71.00
Explanation
2(NH4)2HPO4) → P2O5 2(36+1+31+64) → 62.80264 → 142 ∴ % of P2O5=(142/264)×100=53.78Answer: (c)
Q.16
A molal solution is one that contains 1 mole of a solute in .. [ MP PMT 1999]
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a) 1000g of solvent
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b) one litre of solvent
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c) one litre of the solution
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d) 22.4 litre of the solution
Explanation
Answer: (a)
Q.17
Irrespective of source, pure sample of water always yield 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of ... [ kerala CEE 2002]
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a) conservation of mass
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b) constant composition
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c) multiple proportions
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d) constant proportion
Explanation
Answer: (b)
Q.18
Among FeSO4.7H2O (A), CuSO4.5H2O (B), ZnSO4.7H2O (C), MnSO4.4H2O (D), isomorphous salts are ... [ DPMT 2003]
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a) A and C
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b) A and D
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c) C and B
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d) A anad B
Explanation
Isomorphous substances usually have similar chemical formulas, and the polarizability and ratio of anion and cation radii are generally comparable Ionic radii of Fe2+ is 70 pm and that of Zn2+ is 74pm Answer: (a)
Q.19
One mole of calcium phosphide on reaction with excess of water gives .. [ IIT 1999]
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a) one mole of phosphine
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b) two moles of phosphoric acid
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c) two moles of phosphine
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d) one mole of phosphrous pentoxide
Explanation
Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3 Answer: (c)
Q.20
Which of the following contains maximum number of atoms? [ JIPMER 2000]
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a) 6.023×1021 molecules of CO2
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b) 22.4 L of CO2 at STP
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c) 0.44g of CO2
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d) none of these
Explanation
Answer: (b)
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