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Solutions - Expressing Concentration of Solutions

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A solution is a homogeneous mixture of two or more substances. The component present in the largest quantity is the solvent, and others are solutes.

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Mass Percentage (w/ww/w) represents the mass of a component per 100100 grams of solution. It is commonly used in industrial chemical applications.

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Volume Percentage (v/vv/v) is used for liquid solutes dissolved in liquid solvents, representing the volume of a component per 100100 units of total volume.

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Mass by Volume Percentage (w/vw/v) is often used in medicine and pharmacy, representing the mass of solute dissolved in 100 mL100\text{ mL} of solution.

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Parts per million (ppmppm) is used to express concentration when a solute is present in trace quantities, such as pollutants in water or atmosphere.

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Mole Fraction (xx) is a unitless quantity representing the ratio of moles of a particular component to the total number of moles in the solution. The sum of all mole fractions in a solution is always unity: ∑xi=1\sum x_i = 1.

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Molarity (MM) is the number of moles of solute dissolved in one litre of solution. It is temperature-dependent because volume changes with temperature.

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Molality (mm) is the number of moles of solute per kilogram of solvent. Unlike molarity, molality is independent of temperature as it involves only mass.

📐Formulae

\text{Mass % of a component} = \frac{\text{Mass of component in the solution}}{\text{Total mass of the solution}} \times 100

Mole fraction of component A (xA)=nAnA+nB+nC+...\text{Mole fraction of component A } (x_A) = \frac{n_A}{n_A + n_B + n_C + ...}

Molarity (M)=Moles of soluteVolume of solution in litres\text{Molarity } (M) = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}}

Molality (m)=Moles of soluteMass of solvent in kg\text{Molality } (m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}

Parts per million (ppm)=Number of parts of the componentTotal number of parts of all components×106\text{Parts per million } (ppm) = \frac{\text{Number of parts of the component}}{\text{Total number of parts of all components}} \times 10^6

Relation between M and d:M=% strength×d×10Molar mass of solute (where d is density in g/mL)\text{Relation between } M \text{ and } d: M = \frac{\% \text{ strength} \times d \times 10}{\text{Molar mass of solute}} \text{ (where } d \text{ is density in g/mL)}

💡Examples

Problem 1:

Calculate the molarity of a solution containing 5 g5\text{ g} of NaOHNaOH in 450 mL450\text{ mL} solution.

Solution:

Molar mass of NaOH=23+16+1=40 g mol−1\text{Molar mass of } NaOH = 23 + 16 + 1 = 40\text{ g mol}^{-1} Moles of NaOH=5 g40 g mol−1=0.125 mol\text{Moles of } NaOH = \frac{5\text{ g}}{40\text{ g mol}^{-1}} = 0.125\text{ mol} Volume of solution in litres=4501000=0.450 L\text{Volume of solution in litres} = \frac{450}{1000} = 0.450\text{ L} Molarity=0.125 mol0.450 L=0.278 M\text{Molarity} = \frac{0.125\text{ mol}}{0.450\text{ L}} = 0.278\text{ M}

Explanation:

To find molarity, first calculate the number of moles of the solute (NaOHNaOH) using its molar mass, then divide by the total volume of the solution converted to litres.

Problem 2:

Calculate the molality of 2.5 g2.5\text{ g} of ethanoic acid (CH3COOHCH_3COOH) in 75 g75\text{ g} of benzene.

Solution:

Molar mass of CH3COOH=12×2+1×4+16×2=60 g mol−1\text{Molar mass of } CH_3COOH = 12 \times 2 + 1 \times 4 + 16 \times 2 = 60\text{ g mol}^{-1} Moles of CH3COOH=2.560=0.0417 mol\text{Moles of } CH_3COOH = \frac{2.5}{60} = 0.0417\text{ mol} Mass of benzene (solvent) in kg=751000=0.075 kg\text{Mass of benzene (solvent) in kg} = \frac{75}{1000} = 0.075\text{ kg} Molality (m)=0.0417 mol0.075 kg=0.556 mol kg−1\text{Molality } (m) = \frac{0.0417\text{ mol}}{0.075\text{ kg}} = 0.556\text{ mol kg}^{-1}

Explanation:

Molality is calculated by dividing the moles of solute by the mass of the solvent in kilograms. It is preferred over molarity for experiments where temperature varies.

Problem 3:

A 20%20\% (by mass) solution of KIKI has a density of 1.202 g mL−11.202\text{ g mL}^{-1}. Calculate its molality.

Solution:

20% solution means 20 g of KI in 100 g of solution.20\% \text{ solution means } 20\text{ g of } KI \text{ in } 100\text{ g of solution.} Mass of solvent (water)=100−20=80 g=0.080 kg\text{Mass of solvent (water)} = 100 - 20 = 80\text{ g} = 0.080\text{ kg} Molar mass of KI=39+127=166 g mol−1\text{Molar mass of } KI = 39 + 127 = 166\text{ g mol}^{-1} Moles of KI=20166=0.1205 mol\text{Moles of } KI = \frac{20}{166} = 0.1205\text{ mol} Molality (m)=0.12050.080=1.506 m\text{Molality } (m) = \frac{0.1205}{0.080} = 1.506\text{ m}

Explanation:

Since mass percentage is given, we assume 100 g100\text{ g} of solution. We determine the mass of the solvent by subtraction and use it to find molality.