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NEET Chemistry MCQ
Solutions And Liquid States Mcq Neet Chemistry
Quiz 2
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Q.1
"Relative lowering in vapour pressure of solution containing non-volatile solute is directly proportional to mole fraction of solute." Above statement is : [AFMC 2004]
0%
a) Henry law
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b)Dulong and Petit law
0%
c)Raoult's law
0%
d)Le-Chatelier's principle
Explanation
Raoults law states that relative lowering of vapour pressure is equal to mole fraction of solute .P0 is the pressure of pure solvent , P is pressure of solutionn=moles of solute , N moles of solvent answer: (c)
Q.2
To prepare a solution of concentration of 0.03 g/mL of AgN03, what amount of AgN03 should be added to 60 mL of solution? [AFMC 2005]
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a) 1.8 g
0%
b)0.8 g
0%
c) 0.18 g
0%
d)None of these
Explanation
1ml=0.03 g 60ml=? AgN03=1.8 gAnswer: (a)
Q.3
0.5 M of H2SO4 is diluted from 1 L to 10 L, normality of resulting solution is : [AFMC 2005]
0%
a) 1N
0%
b) 0.1N
0%
c) 10N
0%
d) 11 N
Explanation
Normality of 0.5M of H2SO4 is Normality=Molarity × Basicity=0.5 × 2=1.0 NN1V1=N2V21.0x1=N2 × 10N2=0.1Answer : (b)
Q.4
A 5 molar solution of H2SO4 is diluted from 1 L to 10 L. What is the normality of the solution?[AFMC 2005]
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a)0.25N
0%
b) 1 N
0%
c)2N
0%
d)7N
Explanation
Normality of 5M of H2SO4 is Normality=Molarity × Basicity=5 × 2=10 NN1V1=N2V210.0×1=N2 × 10N2=1Answer : (b)
Q.5
Isotonic solutions have: [AFMC 2006]
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a)same vapour pressure
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b)same osmotic pressure
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c)same boiling point
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d)same temperature
Explanation
Isotonic solutions have same concentration Hence same osmotic pressureAnswer: (b)
Q.6
Methanol and ethanol are miscible in water due to: [AFMC 2007]
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a) covalent character
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b) hydrogen bonding character
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c)oxygen bonding character
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d)none of the above
Explanation
Methanol and ethanol molecules are polar. Oxygen have partial negative charge while oxygen have positive chare . Water is also polar molecule. Hence there is hydrogen bonds between Water and Methanol, Ethanol make them solubleAnswer: (b)
Q.7
Addition of common salt ro a sample of water will ...[ AFMC 1994]
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a) increases its freezing point and increases the boiling point
0%
b) decreases its freezing point and increases the boiling point
0%
c) increases both the boiling point and the freezing point
0%
d) decreases both the boiling and freezing point
Explanation
Answer: (b)
Q.8
n aqueous solution of glucose is 10% in strength. The volume in which 1 g mole of it is dissolved will be [AFMC 2007]
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a) 18L
0%
b) 9L
0%
c) 0.9L
0%
d) 1.8L
Explanation
1 g mole of glucose is molecular mass of glucose=180gSolution is 10% then10g in 100ml180g in ? mlRequired Volume=(180×100) / 10=1800 ml=1.8LAnswer: (d)
Q.9
.01 mole of a non-electrolyte is dissolved in 10g of water. The molality of the solution is [ AFMC 2008]
0%
a)0.1 m
0%
b)0.5 m
0%
c)1.0m
0%
d)0.18m
Explanation
---(1) Here weight of Solute=w Molecular weight Mo Now w/Mo=Number of moles=nThus our formula becomes m: (n×1000)/W ---(2)Given n=0.01Weight of solvent W=10gSubstituting in equation (2) we getM=1mAnswer : (c)
Q.10
hich of the following solutions will exhibit highest boiling point? [AFMC 2008]
0%
a)0.01 M Na2SO4 (aq)
0%
b) 0.01 M KNO3 (aq)
0%
c) 0.015 M urea (aq)
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d) 0.015 M glucose (aq)
Explanation
Elevation in boiling point is a colligative property. More will be the increase in boiling point if number of particle are more On dissolution of.. 0.01M Na2SO4 in one liter number of ions liberated=2 of Na+ + 1 of SO4- , 3 ions from one molecule So concentration of ion will be 0.03 M0.01 M KNO3 in one liter number of ions liberated=1 of K+ + 1 of NO3- 2 ions from one molecule So concentration of ion will be 0.02 M0.015 M urea (aq) do not dissociates hence concentration of particles 0.015M0.015 M glucose (aq) do not dissociates hence concentration of particles 0.015MFrom above maximum concentration of particle is of Na2SO4therefore Na2SO4 will exhibit highest boiling pointAnswer: (a)
Q.11
n pressure cooker food is made faster because [AFMC 2009]
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a) boiling point of water is lowered
0%
b) boiling point of water is increased.
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c)heat is concentrated on food only
0%
d)heat spreads every where.
Explanation
Boiling point of liquid increases with increase in pressureIn pressure cooker boiling point of water increases due to pressure and food is cooked fastAnswer: (b)
Q.12
hich of the following is a colligative property? [AFMC 2009]
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a) Surface tension
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b) Viscosity
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c) Osmotic pressure
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d) Boiling point
Explanation
Increase in boiling point is a colligative property.Osmotic pressure is a colligative propertyAnswer: (c)
Q.13
Relation between partial pressure and mole fraction is stated by [AFMC 2009]
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a)Graham's law
0%
b)Raoult's law
0%
c)Le-Chatelier
0%
d)Avogadro law
Explanation
According to Raoult law for volatile solute and volatile solvent. Partial pressure of Solute is directly proportional to its mole fraction: PA ∝ XA Here PA is partial pressure of solute and XA is mole fraction of soluteAnswer: (b)
Q.14
0.1N 50ml H2SO4 , (N/3) 30 ml HNO3, 0.5N 10ml HCl is mixed and solution is made to 1L. Then normality of resultant solution is [AFMC 2009]
0%
a) N/20
0%
b)N/40
0%
c)N/50
0%
d)N
Explanation
According to law of equivalenceN1V1 + N2V2 + N3V3=NRVRSubstituting given values in above equation0.1×50 + (1/3)(30) + 0.5×10=NR ×1000NR=N/50 Answer: (c)
Q.15
Vapour pressure of benzene at 30°C is 121.8mm When 15g of a non-volatile solute is dissolved in 250 g of benzene, its vapour pressure is decreased to 120.2 mm. The molecular weight of the solute is [AFMC 2009]
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a)35.67 g
0%
b)356.7 g
0%
c) 432.8 g
0%
d) 502.7 g
Explanation
Vapour pressure of solvent (Benzene) Po=121.8 mm Vapour pressure of solution P=120.2 mm Moles of solute n=w/Mo n=15/M0 Moles of solvent N=w/M01 ∴ N=250/78 According to Raoult’s law for nonvolatile solute in a volatile solvent relative decrease in vapour pressure of solution neglecting n in comparison with N We getBy substituting values in above equation Answer: (b)
Q.16
The molal elevation constant for water is 0.What will be the boiling point of 2 molar sucrose solution at 1 atm. pressure? (Assume B.P. of pure water as 100°C) [AFMC 2011]
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a) 101.04°C
0%
b) 100.26°C
0%
c) 100.52°°C
0%
d) 99.74°C
Explanation
Given Kb=0.52 Molarity M=2 Molalty=2 ΔTb=Kb×m=0.52x2=1.04Boiling point of solution=Boiling point of pure solvent + ∆Tb=100+1.04=101.04oCAnswer: (a)
Q.17
At a particular temperature, the vapour pressures of two liquids A and B are respectively 120 and 180 mm of mercury. If 2 moles of A and 3 moles of B are mixed to form an ideal solution, the vapour pressure of the solution at the same temperature will be (in mm of mercury) [AFMC 2011]
0%
a)156
0%
b)145
0%
c)150
0%
d)108
Explanation
According to Raoult’s law for volatile solute and volatile solvent PA=PAo XA here PA is vapour pressure of solute in solution, PAo is vapour pressure of pure solute Now According Dalton’s law of partial pressure P=PA + Pb=PAo XA + PBBo XB Now mole fraction of A,XA=2/5=0.4Now mole fraction of B, XB=3/5=0.6Substituting the values in equation (2) P=120×0.4 + 180×0.6=156 mm of mercuryAnswer: (a)
Q.18
At same temperature, which pair of the following solutions are isotonic solution? (GUJCET 2006)
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a) 0.1M NaCl and 0.1M K2SO4
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b)0.1M Ba(NO3)2 and 0.1M Na2SO4
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c) 0.2M BaCl and 0.2M urea
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d)0.1M urea and 0.1M NaCl
Explanation
For isotonic solution concentration of ion’s or particles must be same in both solutions. Option(a)For 0.1M NaCl, NaCl will dissociate in two ions therefore concentration of ions will be now 0.2 For 0.1M K2SO4, K2SO4 will dissociate in Three ions therefore concentration of ions will be now 0.3 Hence option is not correctSimilarly for option (c)and option (d) here also concentration of ions are different therefore not correct optionOption((b) Foe 0.1MBa(NO3)2, Concentration of ions will be 0.3M For 0.1M Na2SO44, concentration of ions will be 0.3MAnswer(b)
Q.19
Normality of 0.04M H2SO4 is … (GUJCET 2006)
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a)0.08N
0%
b)0.01N
0%
c)0.04N
0%
d)0.02N
Explanation
Since Normality=Molarity × BasicityBasicity of H2SO4 is 2, since it releases to H+ ionsNormality=0.04 × 2=0.08NAnswer: (a)
Q.20
Freezing point of Urea solution is -0.6OC. How much urea is required to be dissolved in 3kg water? (Molecular wt urea=60g/mole, Kf=1.5OCkg/mole) (GUJCET 2007)
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a) 2.4 gram
0%
b) 3.6 gram
0%
c) 6.0 gram
0%
d) 72 gram
Explanation
From the molal freezing point depression constant formula we get(1)Now freezing point given as -0.6OC. Freezing point of pure water is 0OC Hence ∆Tf=0.6OCWeight of solvent (Water) W=3Kg=3000 gMolecular weight of solute M0=60g/mole Kf=1.5oCkg/moleWe have find w=?By substituting above values in equation(1) we gate w=72 gAnswer: (d)
Q.21
Find out the osmotic pressure of 0.25M aqueous solution of urea at 27oC. ( R=0.082 lit atm / mole K, R=1.987 cal) (GUJCET 2007)
0%
a)0.0615 atm
0%
b)61.5 atm
0%
c) 6.15 atm
0%
d)0.615 atm
Explanation
According to formula osmotic pressure π=CRTHere C=Concentration=0.25 M R=Gas constant=0.082 lit atm / mole K T=Absolute temperature=273+27=300K π=Osmotic Pressure Substituting values in above equation we get π=6.15 atmAnswer (c)
Q.22
Find out the normality of a solution, when 9.8 grams of H2SO4 is dissolved in 500 ml solution (GUJCET 2008)
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a)4.0
0%
b)0.8
0%
c)0.2
0%
d)0.4
Explanation
Normality N=(w×Acidity or basicity ×1000)/(Mo×V)Here w is weight of solute=9.8Basicity of H2SO4=2Volume of solution V in ml=500 mlMolecular weight of Solute Mo=98 By substituting values in above equation we getN=0.4Answer: (d)
Q.23
The increase in boiling point of solution containing 0.6 gram urea in 200gram water is 0.5oC. Find the molal elevation constant.(GUJCET 2009)
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a) 1.0 kg/mol
0%
b) 10 K kg mol
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c)10 K g mol-1
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d)10 K kg mol-1
Explanation
Solution: ∆Tb=(1000×Kb×w)/(Mo×W) Here Boiling point elevation constant Kb=?Increase in boiling point ∆Tb=0.5OCWeight of solute w=0.6 gMolecular weight of solute Urea=60 gWeight of solvent V in ml=200 mlOn substituting values in equation we get Kb=10 K kg mol-1Answer: (d)
Q.24
100ml of liquid A was mixed with 25ml of a liquid B to give a non-ideal solution of A-B mixture. The volume of this mixture will be.... [ Haryana CEET 1991]
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a) 75ml
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b) 125ml exact
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c) fluctuating between 75ml and 125ml
0%
d) close to 125 ml but not exceed 125ml
Explanation
Answer: (d)
Q.25
How many grams of NaOH will be required to prepare 500grams solution containing 10% w/w NaOH solution? (GUJCET 2009)
0%
a) 50 gram
0%
b) 5.0 gram
0%
c) 4.0 gram
0%
d) 40.0 gram
Explanation
Mass of NaOH=50 gram Answer : (a)
Q.26
Calculate the normality of 250 ml aqueous solution of H2SO4 having pH=0.00 (GUJCET 2010)
0%
a) 2N
0%
b)1N
0%
c)0.5N
0%
d)0.25N
Explanation
Solution: pH=- log ([H3O+1])pH is 0.00 hence [H3O]=1MSince H2SO4is dibasic one mole of H2SO4 will release 2 moles of H2SO4. There for H2SO4 solution must b 0.5M Now Normality=Molarity × Basicity=0.5 ×2=1.0 N Answer: (b)
Q.27
How many grams of sulphuric acid is to be dissolved to prepare 200 ml aqueous solution having concentration of [H3O+] ions 1M at 25OC temperature [H=1, O=16, S=32 gram/mole](GUJCET 2011)
0%
a) 4.9 gram
0%
b)19.6 gram
0%
c)9.8 gram
0%
d)0.98 gram
Explanation
Concentration of [H3O]=1MSince H2SO4 is dibasic one mole of H2SO4 will release 2 moles of [H3O+] Therefore H2SO4 solution must be 0.5M Here weight of solute w=?Molecular weight of solute Mo=98gVolume of solution in ml V=200Molarity M=0.5By substituting values in above equation we gatew=9.8 gAnswer: (c)
Q.28
The vapour pressure of benzene at 30oC is 121.8mm. By adding 15g of non-volatile solute in 250g of Benzene, its vapour pressure is decreased to 120.2 mm. The molecular weight of solute is... [AIIMS 1997]
0%
a) 156.6g
0%
b) 267.4 g
0%
c) 356.3g
0%
d) 467.4g
Explanation
According to Raoult's law, if solute moles are very less equation ---(1) number of moles of Benzene N = Weight of Benzene/ molecular weight of benzene N = 250/78 = 3.205 moles number of moles of solute n = Weight of solute/ molecular weight of solute (M) n = 15/M Vapour pressure of benzene = po=121.8mm Vapour pressure of solution = p = 120.2 mm Substituting above values in equation --1 we get m = 356.279 Answer: (c)
Q.29
Temperature does not affect: [AIIMS 1997, 2001]
0%
a) Molarity
0%
b) Formality
0%
c) Molality
0%
d) Normality
Explanation
Temperature does not affect molality as it does not depends on volume of solution or solvent Answer: (c)
Q.30
When common salt is dissolved in water
0%
a) Boiling point of solution decreases
0%
b) Melting point of solution increases
0%
c) Boiling point of solution increases
0%
d) Both melting point and boiling point decreases
Explanation
Answer: (c)
0 h : 0 m : 1 s
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