Solutions in Chemistry for CSIR-NET Life Sciences
(Easy Explanation with Exam-Oriented Points)
Understanding solutions is very important for CSIR-NET Life Sciences, especially in Biochemistry, Physical Chemistry, and Molecular Biology experiments. Many questions are directly based on concentration calculations, dilution, ionic strength, and colligative properties.
1️⃣ Percentage and ppm Solutions
๐น What is a Percentage Solution?
Percentage (%) expresses the amount of solute present in 100 parts of solution. There are three common types:
- 1. w/v % (weight/volume): Grams of solute in 100 mL solution.
$$ \text{w/v } \% = \frac{\text{grams of solute}}{\text{mL of solution}} \times 100 $$
๐ Example: 5% NaCl (w/v) = 5 g NaCl in 100 mL solution.
- 2. w/w % (weight/weight): Grams of solute in 100 g solution.
$$ \text{w/w } \% = \frac{\text{grams of solute}}{\text{grams of solution}} \times 100 $$
- 3. v/v % (volume/volume): mL of solute in 100 mL solution. Used for liquids (like ethanol in water).
๐น ppm (Parts Per Million)
ppm is used for very dilute solutions.
$$ \text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6 $$
In aqueous solutions:
$$ 1 \text{ ppm} = 1 \text{ mg/L} $$
๐ Very important for: Environmental toxicology, Heavy metal detection, Trace element analysis.
2️⃣ Molar Solution (Molarity)
๐น What is Molarity (M)?
Molarity = Number of moles of solute per liter of solution.
$$ M = \frac{\text{moles of solute}}{\text{volume in liters}} $$
Since:
$$ \text{Moles} = \frac{\text{Weight (W)}}{\text{Molecular Weight (MW)}} $$
We can write:
$$ M = \frac{W}{MW \times V_{Liters}} $$
๐ Example: Prepare 1M NaCl solution.
Molecular weight of NaCl = 58.5 g/mol.
So, dissolve 58.5 g NaCl in 1 liter of solution.
Molecular weight of NaCl = 58.5 g/mol.
So, dissolve 58.5 g NaCl in 1 liter of solution.
๐ Important CSIR-NET Points:
- Molarity depends on temperature (because volume changes with temperature).
- Used extensively in buffer preparation.
- Very common in Part B and Part C numerical questions.
3️⃣ Normal Solution (Normality)
๐น What is Normality (N)?
Normality = Gram equivalent per liter of solution.
$$ N = \frac{\text{gram equivalent}}{\text{liter of solution}} $$
Gram Equivalent is calculated as:
$$ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} $$
๐ n-factor depends on the reaction:
- Acid → number of H⁺ released
- Base → number of OH⁻ released
- Redox → electrons transferred
๐ Example: H₂SO₄ gives 2 H⁺ (n-factor = 2).
Molecular weight = 98.
Equivalent weight = 98 / 2 = 49.
So, 1N H₂SO₄ = 49 g per liter.
Molecular weight = 98.
Equivalent weight = 98 / 2 = 49.
So, 1N H₂SO₄ = 49 g per liter.
๐ Important Difference
| Feature | Molarity (M) | Normality (N) |
|---|---|---|
| Basis | Based on moles | Based on equivalents |
| Reaction Dependency | Independent of reaction | Depends on reaction (n-factor) |
| Common Use | General chemistry & buffers | Titrations |
4️⃣ Dilution of Solution
Very important for laboratory calculations and experimental design questions.
๐น Dilution Formula:
$$ C_1V_1 = C_2V_2 $$
Where: C₁ = Initial concentration, V₁ = Initial volume, C₂ = Final concentration, V₂ = Final volume.
๐ Example: Prepare 100 mL of 0.1M NaCl from 1M stock.
$$ 1 \times V_1 = 0.1 \times 100 $$ $$ V_1 = 10 \text{ mL} $$ Action: Take 10 mL of stock and add 90 mL of water.
$$ 1 \times V_1 = 0.1 \times 100 $$ $$ V_1 = 10 \text{ mL} $$ Action: Take 10 mL of stock and add 90 mL of water.
5️⃣ Ionic Strength of Solution
Crucial in Biochemistry for understanding protein stability and DNA interactions.
๐น Definition:
Ionic strength (I) measures the total concentration of ions in a solution.
$$ I = \frac{1}{2} \sum C_i Z_i^2 $$
Where: Cแตข = concentration of the ion, Zแตข = charge of the ion.
๐ Example: 0.1 M NaCl
NaCl → Na⁺ + Cl⁻
$$ I = \frac{1}{2} \left[ (0.1 \times 1^2) + (0.1 \times (-1)^2) \right] $$ $$ I = \frac{1}{2} (0.2) = 0.1 $$
NaCl → Na⁺ + Cl⁻
$$ I = \frac{1}{2} \left[ (0.1 \times 1^2) + (0.1 \times (-1)^2) \right] $$ $$ I = \frac{1}{2} (0.2) = 0.1 $$
6️⃣ Colligative Properties of Solution
A very important conceptual topic for CSIR-NET.
๐น What are Colligative Properties?
Properties that depend only on the number of particles (solute), not their nature.
- 1. Relative Lowering of Vapour Pressure: Adding a non-volatile solute decreases vapour pressure.
- 2. Elevation of Boiling Point:
$$ \Delta T_b = K_b \cdot m $$
(where Kb = boiling point constant, m = molality)
- 3. Depression of Freezing Point: Very important in cryopreservation.
$$ \Delta T_f = K_f \cdot m $$
- 4. Osmotic Pressure:
$$ \Pi = MRT $$
(where M = molarity, R = gas constant, T = temperature)
๐ Quick Revision Points
- 1 ppm = 1 mg/L
- Molarity = moles/L
- Normality = equivalents/L
- Dilution: $$ C_1V_1 = C_2V_2 $$
- Ionic strength depends on the square of the charge.
- Colligative properties depend strictly on the number of particles.
CSIR-NET Practice: Solutions
-
Which of the following concentration units is independent of temperature?
A) Molarity
B) Normality
C) Molality
D) Volume percentage
Show Answer
Answer: C. Molality is defined as moles of solute per kg of solvent. Since mass does not change with temperature, molality remains constant. Molarity and Normality depend on volume, which changes with temperature. -
What is the ionic strength of a 0.1 M solution of CaCl2?
A) 0.1 M
B) 0.2 M
C) 0.3 M
D) 0.4 M
Show Answer
Answer: C. CaCl2 dissociates into one Ca2+ (c=0.1, z=2) and two Cl- (c=0.2, z=1).
I = 1/2 Σ cizi2 = 1/2 [ (0.1)(2)2 + (0.2)(-1)2 ] = 1/2 [ 0.4 + 0.2 ] = 0.3 M. -
To prepare 100 mL of 0.5 M buffer from a 5 M stock solution, what volume of the stock solution is required?
A) 5 mL
B) 10 mL
C) 20 mL
D) 50 mL
Show Answer
Answer: B. Using M1V1 = M2V2.
(5 M) × V1 = (0.5 M) × (100 mL).
V1 = 50 / 5 = 10 mL. -
Red blood cells placed in a hypertonic solution will:
A) Swell and burst
B) Shrink due to exosmosis
C) Remain unchanged
D) Divide rapidly
Show Answer
Answer: B. In a hypertonic solution, the concentration of solutes outside the cell is higher than inside. Water will move out of the cell (exosmosis), causing it to shrink or crenate. -
Which of the following properties depends strictly on the number of solute particles in a solution?
A) Refractive index
B) Surface tension
C) Viscosity
D) Osmotic pressure
Show Answer
Answer: D. Osmotic pressure is a colligative property. Colligative properties depend solely on the ratio of the number of solute particles to solvent molecules, not on the identity of the solute. -
What is the Normality of a 2.0 M solution of H2SO4?
A) 1.0 N
B) 2.0 N
C) 4.0 N
D) 8.0 N
Show Answer
Answer: C. Normality = Molarity × n-factor. For H2SO4, it can donate 2 protons (H+), so the n-factor is 2. N = 2.0 M × 2 = 4.0 N. -
The sum of the mole fractions of all components in a solution is always equal to:
A) 0
B) 1
C) 100
D) Depends on the number of components
Show Answer
Answer: B. By definition, a mole fraction represents a part of the whole. The sum of all mole fractions in a given mixture or solution always equals exactly 1. -
If a protein is subjected to "salting out" during purification, what happens at the molecular level?
A) High ionic strength disrupts protein-water interactions, causing precipitation.
B) Low ionic strength causes the protein to unfold.
C) Salt molecules covalently bind to the protein active site.
D) The salt breaks the peptide bonds of the protein.
Show Answer
Answer: A. At high salt concentrations (high ionic strength), salt ions compete with the protein for water molecules. This decreases protein hydration, leading to hydrophobic interactions between protein molecules and their precipitation. -
Which of the following aqueous solutions will have the highest osmotic pressure at 25°C? (Assume complete dissociation)
A) 0.1 M Glucose
B) 0.1 M NaCl
C) 0.1 M CaCl2
D) 0.1 M Urea
Show Answer
Answer: C. Osmotic pressure depends on the number of particles (van 't Hoff factor, i). Glucose and Urea are non-electrolytes (i=1). NaCl gives 2 ions (i=2). CaCl2 gives 3 ions (1 Ca2+ and 2 Cl-, i=3). Highest i = highest osmotic pressure. -
A 1 molar solution is more concentrated than a 1 molal solution because:
A) Mass of the solvent is greater than volume of the solution.
B) The volume of the solution is greater than the volume of the solvent.
C) 1 liter of solution contains less solvent than 1 kg of solvent.
D) Density of the solute is lower.
Show Answer
Answer: C. A 1 M solution has 1 mole of solute in 1 L of total solution (solute + solvent), meaning it contains slightly less than 1 kg of solvent. A 1 m solution has 1 mole of solute in exactly 1 kg of solvent. Therefore, the 1 M solution has less solvent per mole of solute, making it more concentrated.
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