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NEET Chemistry MCQ
Solutions And Liquid States Mcq Neet Chemistry
Quiz 6
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Q.1
An aqueous solution of glucose is 10% in strength. The volume in which one gram mole of it is dissolved will be ... [ AIIMS 1994]
0%
a) 18 litres
0%
b) 9 litre
0%
c) 0.9 litres
0%
d) 1.8 litre
Explanation
10% strength means 100 g in litre or (100/180) moles in one litre∴ volume contain 1 mole or 180g will 1.8 litreAnswer: (d)
Q.2
OSmotic pressure is 0.0821 atm at a temperature of 300K. Find concentration in mole/litre
0%
a) 0.033
0%
b) 0.066
0%
c) 0.33 ×10-2
0%
d) 3
Explanation
use formula π=CRT Answer: (c)
Q.3
The boiling point of a solution of 0.11g of substance in 15g of ether was found to be 0.1°C higher than that of pure ether. The moleculr weight of the substance will be ( Kb=2.16 ) [ Mp PMT 2002]
0%
a) 148
0%
b) 158
0%
c) 168
0%
d) 178
Explanation
Answer: (b)
Q.4
A 500g toothpaste sample has 0.2g fluoride concentration. What is the concentration of fluorine in terms of ppm level? ...[ AIIMS 1994]
0%
a) 250
0%
b) 200
0%
c) 400
0%
d) 1000
Explanation
In 500g=0.2g F- ion In 10+6=400 Answer: (c)
Q.5
The osmotic ptrssure in atmosphere of 10% solution of cane sugar at 69°C .. .. [ AFMC 1991]
0%
a) 724
0%
b) 824
0%
c) 8.21
0%
d) 7.21
Explanation
10% canesugar=(100/342) moles in litre or (100/342) M solution π=CRT=(100/342) ×0.082×(342)=8.21 atmAnswer: (c)
Q.6
Normality of 2M sulphuric acid is ... [ AIIMS 1992]
0%
a) 2N
0%
b) 4 N
0%
c) N/2
0%
d) N/4
Explanation
Normality=Basicity × molarity Basicity of Sulphuric acid is 2Answer: (b)
Q.7
The osmotic pressure of a solution at 0°C is 4 atm. What will be its osmotic pressure 546K under similar condition
0%
a) 4 at,
0%
b) 2 atm
0%
c) 8 atm
0%
d) 1 atm
Explanation
pressure ∝ temperature temperature is doubled pressure will be doubledAnswer: (c)
Q.8
1000 g of a 4% solution of a salt contains: .. [ AIIMS 1992]
0%
a) 4 g salt
0%
b) 4 moles of salt
0%
c) 40g of solvent
0%
d) 40g of salt
Explanation
Answer: (d)
Q.9
The rise in the boiling point of a solution containing 1.8g of glucose in 100g of solvent is 0.1°C. The molal elevation constant of the liquid is ...[ Manipal PMT 2001]
0%
a) 0.01 K/m
0%
b) 0.1 K/m
0%
c) 1 K/m
0%
d) 10 K/m
Explanation
Molecular wt. of glucose=180g; 1.8g of glucose in 100g of solvent molality=0.1 ΔTb=Kb m or Kb=1 K/mAnswer: (c)
Q.10
In a mixture A and B components show negative deviation as .... [ AIEEE 2002]
0%
a) ΔVmix > 0
0%
b) ΔHmix < 0
0%
c) A-B interaction is weaker than A-A and B-B interaction
0%
d) none of the above reason correct
Explanation
Answer: (b)
Q.11
Formation of a solution from two components can be considered asi) pure solvent → separated solvent molecules, ΔH1ii) pure solute → separated solute molecules, ΔH2iii) separated solvent and solute molecules → solution, ΔH3Solution so formed will be ideal is ... [ CBSE Med. 2003]
0%
a) ΔHsoln=ΔH1 + ΔH2 + ΔH3
0%
b) ΔHsoln=ΔH1 + ΔH2 - ΔH3
0%
c) ΔHsoln=ΔH1 - ΔH2 - ΔH3
0%
d) ΔHsoln=ΔH3 - ΔH1 - ΔH2
Explanation
if net ΔHsoln is the sum of three steps, this mens that solute-solvent interactions are similar to solvent-solvent and solute-solute interactions. Answer: (a)
Q.12
Which one of the statements given below concerning properties of solutions describes a colligative effect? [ AIIMS 2003]
0%
a) Boiling point of pure water decreases by the addition of ethanol
0%
b) vapour pressure of pure water decreases by the addition of nitric acid
0%
c) Vapour pressure of pure benzene decreases by the addition of naphthalene
0%
d) boiling point of pure benzene increases by addition of toluene
Explanation
Answer: (c)
Q.13
The molarity of solution prepared by adding 7.1g of Na2SO4 ( formula weight 142u) to enough water to make 100ml volume is .. [ MP PET 1993]
0%
a) 2.0 M
0%
b) 1.0 M
0%
c) 0.5 M
0%
d) 0.05 M
Explanation
Answer: (c)
Q.14
The depression in freezing point for 1M urea, 1M glucose and 1M NaCl are in the ratio ... [ Haryana CEET 2000]
0%
a) 1:2:3
0%
b) 3:2:2
0%
c) 1:1:2
0%
d) none of these
Explanation
1 M urea and 1M glucose will have same depression in freezing point while it will be double for 1M NaCl Answer: (c)
Q.15
If liquid A and B form ideal solution ... [ AIEEE 2003]
0%
a) the entropy of mixing is zero
0%
b) the free energy of mixing is zero
0%
c) the free energy as well as the entropy of mixing are each zero
0%
d) the enthalpy of mixing is zero
Explanation
Answer: (d)
Q.16
The solubility of a gas in water depends on ... [ MP PET 2002]
0%
a) Nature of gas
0%
b) Temperature
0%
c) Pressure of the gas
0%
d) all of the above
Explanation
Answer: (c)
Q.17
At room temperature, the mole fraction of a solute is 0.25 and the vapour pressure of a solvent is 0.80 atm. Then the lowering of vapour pressure is .. [ Taminadu CET 2002]
0%
a) 0.75
0%
b) 0.60
0%
c) 0.20
0%
d) 0.80
Explanation
Answer: (c)
Q.18
Which one of the following aqueous solutions will have the lowest freezing point? [ kerala CEE 2001]
0%
a) 0.1 molal solution of urea
0%
b) 0.1 molal solution of sucrose
0%
c) 0.1 molal solution of sodium chloride
0%
d) 0.1 molal solution of calcium chloride
Explanation
Concentration of particles in 0.1m solution of calcium chloride is 0.3m. Which is maximum. Hence depression in freezing point is maximum Answer: (d)
Q.19
During depression of freezing point in a solution, the following are in equilibrium.. [ IIT 2003]
0%
a) liquid solvent, solid solvent
0%
b) liquid solvent, solid solute
0%
c) liquid solute, solid solid solute
0%
d) liquid solute, solid solvent
Explanation
Answer: (a)
Q.20
ISotonic solutions have .. [ DPMT 2000]
0%
a) same boiling point
0%
b) same vapour pressure
0%
c) same melting point
0%
d) same osmotic pressure
Explanation
Answer: (d)
Q.21
The experimental molecular weight of an electrolyte will always be less than its calculated value because the value of van't Hoff factor 'i' is value because the value of van't Hoff factor 'i' is ... [ PM PMT 1993]
0%
a) less than 1
0%
b) greater than 1
0%
c) equivalent to 1
0%
d) zero
Explanation
Answer: (b)
Q.22
The vapour pressure of a solution ( P) and the vapour pressure of the solvent (Po) are related to each other as ... [ Tamilnadu CET 2001]
0%
a) P=Pox2
0%
b) Po=Px1
0%
c) P=Pox1
0%
d) Po=Px2
Explanation
Answer: (c)
Q.23
A solution contains non-volatile solute of molecular mass MWhich of the following can be used to calculate the molecular mass of solution in terms of osmotic pressure? ... [ CBSE PMT 2002]m2=mass of solute, V=volume of solution, π=osmotic pressure
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Answer: (b)
Q.24
Vapour pressure of CCl4 at 25°C is 143mm Hg ,0.5g of non-volatile solute ( mol wt 65 ) is dissolved in 100ml of CClFind the vapour pressure of the solution ( Density of CCl4=1.58 g/cm3) [ CBSE PMT 1996]
0%
a) 141.93 mm
0%
b) 84.39 mm
0%
c) 199.34 mm
0%
d) 143.99 mm
Explanation
Answer: (a)
Q.25
To 5.85g of NaCl , one kg of water is added to prepare a solution. What is the strength of NaCl in this solution? ( molecular weight of NaCl=58.5) [ CPMT 1990]
0%
a) 0.1 Normal
0%
b) 0.1 Molal
0%
c) 0.1 Molar
0%
d) 0.1 Formal
Explanation
Answer: (b)
Q.26
The vapour pressure of solvent decreases by 10mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.What should be the mole fraction of the solvent if the decrease in vapour pressure is to be 20mm of mercury? [ CBSE PMT 1998]
0%
a) 0.8
0%
b) 0.6
0%
c) 0.4
0%
d) 0.4
Explanation
ΔP=Po × x2 10=Po × 0.2Po=50 mm Again20=Po × x2 20=50 × x2 x2=0.4 x1=1.0 - 0.4=0.6 Answer: (b)
Q.27
12g H2SO4 was dissolved in water. The volume of the solution was made 1200 ml. The normality of the solution is ... [ NCERT 1980]
0%
a) 0.5 N
0%
b) 0.102 N
0%
c) 0.112 N
0%
d) 0.204 N
Explanation
Answer: (d)
Q.28
When the solute is present in trace quantities the following expression is used ... [ kerala MEE 2002]
0%
a) gram per million
0%
b) milligram percent
0%
c) microgram percent
0%
d) parts per million
Explanation
Answer: (d)
Q.29
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) .. [ kerala CEE 2001]
0%
a) 156
0%
b) 145
0%
c) 150
0%
d) 48
Explanation
mole fraction of A=xA=2/5=0.4 mole fraction of B xB=3/5=0.6PA=PAo ×xA=0.4 ×120=48 mmPB=PBo ×xB=0.6 ×180=108 mmPtotal=48 +108=156 mmAnswer: (a)
Q.30
Identify the mixture that shows positive deviations from Raoult's law ... [ kerala CEE 2002]
0%
a) CHCl3 + (CH3)2CO
0%
b) (CH3)2CO + C2H5NH2
0%
c) (CH3)2CO + CS2
0%
d) C6H5N + CH3COOH
Explanation
Answer: (c)
0 h : 0 m : 1 s
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