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NEET Chemistry MCQ
Chemical Arithmetic And Volumetric Analysis Mcq Neet Chemistry
Quiz 2
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Q.1
Assuming fully decomposed, the volume of CO2 released at STP on heating 9.85g of BaCO3 ( Atomic mass, Ba=137 ) will be... [ CBSE PMT 2000]
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a) 0.84L
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b) 2.24L
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c) 4.06L
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d) 1.12L
Explanation
BaCO3 → BaO + CO2 Molecular wt of BaCO3=1979.85 g=0.05 mole which will give 0.05 mole of CO2at STP 1 mole=22.4 litreat STP 0.05 mole=1.12LAnswer: (d)
Q.2
A mixture of sand and iodine can be separated by .. . [ kerala CEE 2002]
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a) crystallisation
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b) submilation
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c) distillation
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d) fraction
Explanation
Answer: (b)
Q.3
Among the following pairs of compounds, the one that illustrates the law of multiple proportion is .. [ kerala MEE 2001]
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a) NH3 and NCl3
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b) H2S and SO2
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c) CuO and Cu2O
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d) CS2 and FeSO4
Explanation
Answer: (c)
Q.4
A 100 ml solution of 0.1N HCl was titrated with 0.2N NaOH solution. The titration was discontinued after adding 30ml of NaOH solution. The remaining titration was completed by adding 0.25N KOH solution. The volume of KOH required for completing the titration is ... [ DCE 1999]
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a) 70ml
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b) 32ml
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c) 35ml
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d) 16ml
Explanation
Step I miliequivalent of HCl=0.1×100=10miliequivalent of NaOH=0.2 × 30=6miliequivalent of HCl remains=10 -6=4Thus 4 milimole of KOH required 4=0.25 × VV=16 mlAnswer: (c)
Q.5
A compound contains atoms of three elements in A,B and C. If oxidation number of A is +2, B is +5 and that of C is -2, the possible formula of the compound is
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a) A3(BC4)2
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b) A3(B4C)2
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c)ABC2
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d) A2(BC3)2
Explanation
total oxidation number should become zero for option (a)A=3×(+2)=+6B==2 × (+5)=+10C=2×4×(-2)=-16Total of above is zero; option (a) is correctAnswer: (a)
Q.6
7.5g of gas occupy 5.6 litre of volume at STP. The gas is ... [ EAMCET 2001]
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a) NO
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b) N2O
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c) CO
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d) CO2
Explanation
22.4 L at STP=1 mole 5.6 L at STP=0.25 molegiven wt of 0.25 mole=7.5∴ wt. of 1 mole of gas=30g if of gas NOAnswer: (a)
Q.7
For preparing 0.1N solution of a compound from its impure sample of which the percentage purity is known, the weight of the substance required will be ... [ MP PET 1996]
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a) more than the theoretical weight
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b) less than the theoretical weight
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c) same as the theoretical weight
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d) none of the above
Explanation
Answer: (a)
Q.8
10 dm3 of N2 gas and 10 dm3 of gas X at the same temperature contain the same number of molecules. The gas is ... [ KCET 2001]
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a) CO
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b) CO2
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c) H2
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d) NO
Explanation
Number of molecules of N2 and X should be same. This can be so if X has the same molecular weight, This is since there is no mention about same pressure Answer: (a)
Q.9
The simple formula of compound containing 50% of element X ( atomic mass 10) and 50% of element Y ( atomic mass 20) is ... [ Roorkee 1994]
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a) XY
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b) X2Y
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c) XY3
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d) X2Y3
Explanation
Mass ratio 50 : 50 Number of atom ration50/10 : 50/20Or2 : 1Formula X2YAnswer: (b)
Q.10
The number of water molecules present in a drop of water ( volume 0.0018ml) at room temperature is ... [ DCE 2000]
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a) 6.023 × 1019
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b) 1.084×1018
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c) 4.84×1017
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d) 6.023×1023
Explanation
18 ml=18g=1 mole of water 0.0018 ml=0.0018 g=0.0001 mole of water1 mole=6.023×1023 number of molecules0.0001 mole=6.023×1019 number of molecules Answer: (a)
Q.11
The set of numerical coefficients that balances the equationK2CrO4 + HCl → K2Cr2O7 + kCl + H2O is kerala CEE 2001
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a) 1, 1, 2, 2, 1
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b) 2, 2, 1, 1, 1
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c) 2, 1, 1, 2, 1
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d) 2, 2, 1, 2, 1
Explanation
Answer: (d)
Q.12
How much of NaOH is required to neutralise 1500 cm3 of 0.1N HCl? ( Na=23 ) [ KCET 2001]
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a) 40g
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b) 4g
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c) 6g
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d) 60g
Explanation
1500 cm3 of 0.1 N HCl=1500 ×0.1=150 miliequivalent or 0.15g equivalent It will be neutralized by NaOH of 0.15g equivalent=0.15× mol.wt=0.15 × 40=6gAnswer: (c)
Q.13
Number of 'g' of oxygen in 32.2g of Na2SO4 . 10H2O is .. [ Haryana PMT 2000]
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a) 20.8
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b) 22.4
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c) 2.24
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d) 2.08
Explanation
1 mole of Na2SO4 . 10H2O=14 g atom of oxygen Molecular weight of Na2SO4 . 10H2O=322 g contains 224 g of oxygen∴ 32.2 contains 22.4 gAnswer: (b)
Q.14
A certain amount of a metal whose equivalent mass is 28 displaces 0.7 L of H2 at S.T.P. from an acid hence mass of the element is
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a) 1.75 g
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b) 0.875 g
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c) 3.50 g
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d) 7.00 g
Explanation
Equivalent Weight of metal displaces = 11.2 L of H2 gas 11.2L of H2 displaced by 28g of metal × 0.7Lof H2 is displaced by 1.75 g of metal Answer (a)
Q.15
ssertion: In the reaction 2NaOH + H3 PO4 → Na2 HPO4 + 2H2 O equivalent weight of H3PO4 is M/2, where M is its molecular weight. Reason: Equivalent weight =Molecular weight/ n -factor 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) Assertionis correct but Reason is wrong.
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d) Both Assertion and Reason are incorrect.
Explanation
As two Hydrogen of H3PO4 is placed by Na. Thus equivalent weight of H3PO4 is M/2 And reason is standard formula Answer : (a)
Q.16
Assertion: One mole of sucrose reacts completely with oxygen produces 268.8 litre of carbon dioxide at STP. Reason: Amount of oxygen required for reaction is 268.8 litre.
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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 : (b)
Q.17
Assertion: When 4 moles of H2 reacts with 2 moles of O2, then 4 moles of water is formed. Reason: O2 will act as limiting reagent.
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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 : (c)
Q.18
What is the [OH-] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 MBa(OH)2?
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a) 0.12 M
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b) 0.10 M
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c) 0.40 M
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d) 0.0050 M
Explanation
Reaction Ba(OH)2 + 2HCl2 → BaCl2 + 2H2O Miliequivalnce of HCl = 20×0.05 = 1 Miliequivalnce of Ba(OH)2 = 2×30×0.1= 6 ( as charge on Ba =2 or Number of OH =2) Thus 1 miliequivalce of HCl will be neutralized And 5 miliequivalence of Ba(OH)2 will remain unreacted But total volume =20+30 =50 Thus OH- concentration M M = 5/50=0.1M Answer : (b)
Q.19
An organic compound containing C, H and N gave the following analysis C = 40%, H = 13.33%,N = 46.67%. Its empirical formula would be
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a) CH4N
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b) CH5N
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c) C2H7N2
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d) C2H7N
Explanation
Number of carbon atoms =40/12 Number of H atoms = 33.3/1 Number of Nitrogen atoms =46.67/14 From above number of Nitrogen and Carbon atom same Option(a) correct Answer : (a)
Q.20
4 g of hydrogen reacts with 16 g of oxygen to form water. The mass of water formed is
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a) 24 g
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b) 36 g
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c) 22.5 g
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d) 18 g
Explanation
2H2 + O2 → 2 H2O Form above equation Two moles of hydrogen reacts with one mole of Oxygen to give two moles of water Or 4 g of hydrogen reacts with 32 g of oxygen to produce 36g of water It is given oxygen is 16g Now 32g oxygen reacts with 4g of hydrogen ∴ 16 g oxygen reacts with = 16×4/32 = 2.0 g of Hydrogen Thus water produced = 16+2.0 = 18g Answer : (d)
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