Solutions: Concentration Terms (Part-1)

• Yakeen 2.0 For NEET 2025, Lecture-01 Physical Chemistry, Solutions, by Sudhanshu Sir.

Topics

• Concentration Terms (Part-1).

Solutions

• Solutions can expect a minimum of two questions in the NEET-2025 exam.
• Topics covered include:
  • Concentration Terms A
  • Solubility B: covers solid and gas in liquid, liquid in liquid types of solutions, and azeotropic mixture.
  • Colligative Properties C: includes RLVP, ΔTb, ΔTf, and π.
  • Van't Hoff Factor D.

Concentration Terms

• Terms based on percentage includes w/w, v/v, w/v, and v/w.
• Fraction Terms includes mass fraction, volume fraction, and mole fraction.
• Parts per million (ppm) are defined with respect to 10^6
• Parts per billion (ppb) are defined with respect to 10^9
• Molarity (M).
• Molality (m).
• Strength (S).
• Formality (F).
• Normality (N).

Percentage Terms

• w/w or mass percent:
  • (w/w)A = (WA / (WA + WB)) * 100
  • 10% (w/w) NaCl aqueous solution: 10g NaCl is present in 100g solution
• v/v or volume percent:
  • (v/v)A = (VA / (VA + VB)) * 100
  • 10% (v/v) NaCl aqueous solution: 10ml NaCl is present in 100 ml solution
• (w/v) or mass by volume:
  • (w/v)A = (WA / Vtotal) * 100
  • 10% (w/v) NaCl aqueous solution: 10g NaCl is present in 100 ml solution
• v/w or volume by mass:
  • (v/w)A = (VA / (WA + WB)) * 100
  • 10% (v/w) NaCl aqueous solution: 10ml NaCl is present in 100g solution
• Aqueous solution is a solution made in water.
• Alcoholic solution is a solution made in alcohol (C2H5OH).
• 5% (w/w) HCl aqueous solution has 5g HCl present in 100g solution.
• 5% (v/v) HCl aqueous solution has 5ml HCl present in 100ml solution.
• 5% (v/w) HCl aqueous solution has 5ml HCl present in 100 g solution.
• 5% (w/v) HCl aqueous solution has 5g HCl present in 100ml solution.

Binary Solution

• Binary solution contains only two components mixed homogeneously
• Solute: present in less quantity.
  • Mass is less.
  • If mass is equal, no. of moles is less.
  • In a liquid-liquid solution, volume is less.
• Solvent: present in high quantity.
  • Mass is high.
  • If mass is equal, no. of moles is high.
  • In a liquid-liquid solution, volume is high.
• Ternary solution: consists of three components
• Quaternary solution: consists of four components, this is considered out of syllabus.

Fraction Terms

• Values range from 0 to 1
• Mass fraction equals (w2 / (w1+w2))

Volume Fraction

• The volume fraction for component 2 is V2 / (V1 + V2).
• The volume fraction for component 1 is V1 / (V1 + V2).

Mole Fraction

• Represents mole fraction as "x" (pronounced as "Chi").
• Mole fraction calculation: x2 = n2 / (n1 + n2)
• Mole fraction calculation: x1 = n1 / (n1 + n2)
• The sum of mole fractions always equals 1.

Parts per Million (ppm)

• 1 million = 10^6
• (ppm)A = (WA / Wtotal) * 10^6
• (ppm)A = (VA / Vtotal) * 10^6
• 200 ppm F- ion in water corresponds to 200g F- ion in 10^6 gram of water (w/w) or 200 ml F- ion in 10^6 ml of H2O (v/v).

Parts per Billion (ppb)

• 1 billion = 10^9
• (ppb)A = (WA / Wtotal) * 10^9
• (ppb)A = (VA / Vtotal) * 10^9
• 100 ppb Ca+2 ion in water corresponds to 100g Ca+2 ion in 10^9 g of water (w/w) or 100ml Ca+2 ion in 10^9 ml H2O.

Example Problems With Solutions

• To calculate the volume by volume percentage of a solution of 15 ml of alcohol in 60 ml of water: (15 / (15+60)) * 100 = 20%.
• To calculate the mass by volume percentage when 40 g of sugar is dissolved in 600 mL of sugar solution: (40 / 600) * 100 = 6.66%.
• To determine the volume/volume percent solution made by containing 25 mL of ethanol with enough water to produce 200 mL: (25 / 200) * 100 = 12.5%.
• Given 5 × 10^-3 kg of urea dissolved in 2 × 10^-3 kg of water, the percent by mass of urea is: (5 * 10^-3 / (7 * 10^-3)) * 100 = 71.42%.
• If adding 5 g of a substance x to 18 g of water, then the mass percentage of the solute is: (5 / 23) * 100 = 21.74%.
• If 0.025 g of KCl is dissolved in 100 grams of water, then the concentration in ppm is: (0.025 / 100) * 10^6 = 250 ppm.
• For a 500 gm toothpaste sample with 0.2 g fluoride concentration, the concentration of fluoride ions in terms of ppm is: (0.2 / 500) * 10^6 = 400.
• The sum of mole fractions of all the components in a solution adds up to unity
• A solution is prepared by adding 360g of glucose to 864g of water, then the mole fraction of glucose is calculated as approximately 0.04
• 360/180 / (360/180) + (864/18) = 2 / 50 *

Volume Percentage and Molarity

• To calculate concentration with 2.5g potassium chloride to 50ml solution, the mass by volume is 5%
• (2,5 / 50 ) * 100*
• To calculate volume percentage with 75 ml of solute to 155ml volume. percentage = 32.6
• (75 / (75 + 155)) * 100*
• To calculate amount of water need to add to methanol and mole fraction as 0.25 = 27g

Molarity

• Represented as M, or capital M.
• Molarity is defined as the number of moles of solute present in 1 L / 1000 ml of solution.
• Molarity = nsolute (mol) / Vsolution (L), where Unit of Molarity = mol/L or Molar or M
• 1 litre = 1000 ml = 1/1000m^3 = 1000 cm^3 = 1 dm^3 or c.c
• The formulas: Molarity = (n2 / V(L)) = (n2 / V(ml)) * 1000 = (W2 / M2) * (1000 / V(ml)), where W2 is the given mass of solute and M2 is the molar mass of solute.
• 2 M NaCl aqueous solution means of 2 moles of NaCl is present in 1000 ml solution.
• 3 M H2SO4 aqueous solution means 3 moles of H2SO4 is present in 1L/1000ml solution.
• M = (n2 / V(L)) = (n2 / V(ml)) * 1000.
• nsolute = M * V(L).
• nsolute = M * V(ml) / 1000 = M * V(ml) * 10^-3.
• The number of millimoles of solute = M * V(ml).

Example Calculations

• If 1.8 gm glucose is present in 200 ml of solutions, its molarity: (1.8 / 180) * (1000 / 200) = 0.05 M
• A given solution of NaOH contains 4.00 g of NaOH per litre of solution, thus the molarity is: (4 / 40) * (1000/1000) = 0.1
• With 6.022 x 10^20 molecules of acetic acid present in 500 ml of its solution, the concentration of solution: (10^-3)/(1/2) = 2x10^-3 mol/l = 0.002 M
• What volume of a 0.8 M solution contains 100 millimoles of the solute. (100/0.8 = 125ml
• The molarity of the solution containing 2.8% (w/v) solution of KOH equals 0.5 M and given that 2.8g KOH is present in 100 ml solution.

Temperature

• Temperature dependent: involves the volume of liquid. Includes molarity since Molarity = moles/volume and (w/v) = (Wsolute *100) / Vtotal
• Temperature independent: does not involve the volume of liquid includes Molality and W/W = (Wsolute / Wtotal )*100