A solution containing 5 moles of solute in 2 liters of solution has molarity
A 1.0 M
B 2.0 M
C 2.5 M
D 5.0 M
Molarity = moles / volume (L) = 5 / 2 = 2.5 M.
A solution containing 2 moles of solute in 1 kg of solvent has molality
A 1 m
B 2 m
C 0.5 m
D 4 m
Molality = moles of solute / kg of solvent = 2 / 1 = 2 m.
Which concentration term changes with temperature
A molality
B mole fraction
C molarity
D mass percent
Molarity depends on volume, which changes with temperature.
The sum of mole fractions of all components in a solution is
A zero
B less than one
C greater than one
D equal to one
Mole fraction is defined such that total equals 1.
Which of the following is an example of liquid in liquid solution
A fog
B milk
C ethanol in water
D smoke
Both ethanol and water are liquids forming a true solution.
Which unit is commonly used to express concentration of pollutants in air
A molarity
B molality
C ppm
D mole fraction
ppm is suitable for very low concentrations.
If mole fraction of solute is 0.2, mole fraction of solvent is
A 0.2
B 0.8
C 1.2
D 2.0
Sum of mole fractions = 1.
Normality is defined as
A moles per liter
B equivalents per liter
C grams per liter
D moles per kg
Normality depends on reaction and equivalent weight.
Which concentration unit is reaction-dependent
A molarity
B molality
C normality
D mole fraction
Equivalent depends on nature of reaction.
A 10% (w/w) solution contains
A 10 g solute in 100 g solution
B 10 g solute in 100 g solvent
C 10 g solute in 1 L solution
D 100 g solute in 10 g solution
Weight percent is defined per 100 g of solution.
Vapor pressure of a solvent decreases on addition of non-volatile solute because
A solute increases surface area
B mole fraction of solvent decreases
C solute evaporates
D temperature increases
According to Raoult’s law, P = XₛP°.
If mole fraction of solvent is 0.9 and vapor pressure of pure solvent is 100 mmHg, vapor pressure of solution is
A 10 mmHg
B 50 mmHg
C 90 mmHg
D 110 mmHg
P = XₛP° = 0.9 × 100 = 90 mmHg.
Relative lowering of vapor pressure depends on
A nature of solvent
B temperature only
C number of solute particles
D pressure only
It is a colligative property.
An ideal solution shows
A positive deviation
B negative deviation
C no deviation from Raoult’s law
D only vapor pressure lowering
Ideal solutions strictly obey Raoult’s law.
Which solution shows positive deviation from Raoult’s law
A acetone + chloroform
B ethanol + acetone
C HNO₃ + water
D HCl + water
Weaker A–B interactions cause positive deviation.
Negative deviation from Raoult’s law occurs when
A A–B attraction < A–A and B–B
B A–B attraction > A–A and B–B
C vapor pressure increases
D solution becomes ideal
Stronger interactions lower vapor pressure more than ideal.
Maximum boiling azeotrope is formed by solutions showing
A positive deviation
B negative deviation
C ideal behavior
D no vapor pressure
Strong interactions raise boiling point.
Minimum boiling azeotrope is formed by solutions showing
A negative deviation
B positive deviation
C ideal behavior
D no deviation
Weaker interactions increase vapor pressure.
Azeotropes cannot be separated by
A crystallization
B simple distillation
C fractional distillation
D filtration
Composition of vapor equals liquid at azeotrope.
Raoult’s law is applicable to
A ideal solutions only
B non-ideal solutions only
C colloidal solutions
D electrolytic solutions only
Raoult’s law holds exactly for ideal solutions.
Boiling point elevation is independent of
A number of solute particles
B nature of solute
C molality
D van’t Hoff factor
Colligative properties depend only on particle count.
A 1 m solution of urea has ΔTb = 0.52 K. The ebullioscopic constant of solvent is
A 0.26
B 0.52
C 1.04
D 2.0
ΔTb = Kb m → Kb = 0.52 / 1 = 0.52.
Freezing point depression is greater when
A solvent has higher Kf
B solute has higher molar mass
C solute particles are fewer
D solution is dilute
ΔTf = Kf m → larger Kf gives more depression.
Which property is used for molar mass determination of polymers
A ΔTb
B ΔTf
C vapor pressure lowering
D osmotic pressure
Osmotic pressure is measurable for very dilute polymer solutions.
Van’t Hoff factor for K₂SO₄ in dilute solution is approximately
A 1
B 2
C 3
D 4
K₂SO₄ → 2K⁺ + SO₄²⁻ → 3 ions.
If van’t Hoff factor is less than 1, solute undergoes
A dissociation
B association
C ionization
D hydrolysis
Association reduces number of particles.
Osmotic pressure depends on
A molarity
B molality
C mass percent
D mole fraction
π = MRT → depends on molarity.
Which solution will have highest osmotic pressure
A 0.1 M glucose
B 0.1 M NaCl
C 0.05 M Na₂SO₄
D 0.1 M urea
NaCl dissociates into two ions → higher particle count.
Reverse osmosis requires pressure greater than
A vapor pressure
B osmotic pressure
C atmospheric pressure
D surface tension
External pressure must exceed osmotic pressure.
Isotonic solutions have same
A pH
B boiling point
C osmotic pressure
D freezing point
Isotonic means equal osmotic pressure.
Adsorption is always
A endothermic
B exothermic
C temperature independent
D irreversible
Energy is released when molecules stick to surface.
Physical adsorption decreases with increase in
A pressure
B surface area
C temperature
D gas concentration
Weak van der Waals forces break at high temperature.
Chemisorption usually increases with increase in
A pressure only
B temperature
C surface area only
D gas volume
Activation energy is required for chemical bond formation.
Which adsorption is multilayer
A chemisorption
B physical adsorption
C ionic adsorption
D selective adsorption
Weak forces allow multiple layers.
Langmuir adsorption isotherm assumes
A multilayer adsorption
B uniform surface
C temperature variation
D gas solubility
Langmuir model assumes identical adsorption sites.
Promoters increase activity of catalyst by
A blocking surface
B increasing surface area
C poisoning catalyst
D changing equilibrium
Promoters enhance catalytic efficiency.
A negative catalyst
A speeds up reaction
B slows down reaction
C changes equilibrium
D increases ΔG
Negative catalysts (inhibitors) reduce reaction rate.
Enzyme catalysis is highly
A non-specific
B temperature independent
C specific
D irreversible
Enzymes act on specific substrates.
Which is NOT a colloid
A milk
B fog
C sugar solution
D smoke
Sugar solution is a true solution.
Which colloid is solid in gas
A sol
B aerosol
C foam
D gel
Smoke is solid dispersed in gas.
Gold sol is stabilized due to
A Brownian motion only
B electrical charge on particles
C large particle size
D gravity
Charge causes repulsion preventing coagulation.
Which method purifies colloids
A coagulation
B dialysis
C precipitation
D filtration
Dialysis removes electrolytes.
Coagulating power increases with
A decreasing ion charge
B increasing ion charge
C decreasing concentration
D temperature decrease
Hardy–Schulze rule.
Which ion will coagulate negatively charged sol fastest
A Na⁺
B Ca²⁺
C Al³⁺
D K⁺
Higher valency → stronger coagulation.
Zeta potential is related to
A viscosity
B surface charge
C particle size
D density
Zeta potential measures stability of colloids.
Which colloid shows Tyndall effect
A NaCl solution
B sugar solution
C starch sol
D copper sulfate solution
Colloids scatter light.
Emulsifying agents stabilize emulsions by
A decreasing viscosity
B reducing surface tension
C increasing particle size
D causing coagulation
They prevent separation of liquids.
Soap stabilizes oil-water emulsion by
A adsorption
B micelle formation
C precipitation
D osmosis
Soap forms micelles around oil droplets.
Which colloid is lyophobic
A starch sol
B gelatin sol
C gold sol
D gum sol
Metal sols have little affinity for medium and are unstable.
Protective colloids increase stability by
A increasing temperature
B forming adsorption layer
C decreasing surface area
D removing charge
Protective layer prevents aggregation.