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The Amazing World of Solutes, Solvents, and Solutions - Determination of density

Grade 8CBSE

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

🔑Concepts

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A solution is a homogeneous mixture where a solute (the substance being dissolved) is uniformly distributed within a solvent (the substance that does the dissolving).

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Density is a characteristic physical property of a substance defined as its mass per unit volume. It is mathematically represented as ρ\rho (rho) or DD.

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The density of a solution depends on the concentration of the solute. Generally, adding more solute to a solvent increases the total mass of the solution more significantly than the volume, thereby increasing the overall density.

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The standard SI unit for density is kg/m3kg/m^3, but in laboratory settings, g/cm3g/cm^3 or g/mLg/mL is more commonly used, where 1 cm3=1 mL1\text{ cm}^3 = 1\text{ mL}.

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To determine the density of a liquid solution, we first find the mass of the empty beaker (m1m_1), then the mass of the beaker with the solution (m2m_2). The mass of the solution is calculated as m=m2−m1m = m_2 - m_1.

📐Formulae

Density(ρ)=Mass(m)Volume(V)\text{Density} (\rho) = \frac{\text{Mass} (m)}{\text{Volume} (V)}

msolution=mbeaker + solution−mempty beakerm_{\text{solution}} = m_{\text{beaker + solution}} - m_{\text{empty beaker}}

Vobject=Vfinal−VinitialV_{\text{object}} = V_{\text{final}} - V_{\text{initial}}

ρsolution=msolute+msolventVsolution\rho_{\text{solution}} = \frac{m_{\text{solute}} + m_{\text{solvent}}}{V_{\text{solution}}}

💡Examples

Problem 1:

A student wants to find the density of a salt solution. The mass of an empty graduated cylinder is 110 g110\text{ g}. After pouring 50 mL50\text{ mL} of the salt solution into it, the total mass becomes 165 g165\text{ g}. Calculate the density of the solution.

Solution:

  1. Find the mass of the solution: 165−11055\begin{array}{r} 165 \\ - 110 \\ \hline 55 \end{array} Mass (mm) = 55 g55\text{ g}.
  2. Given Volume (VV) = 50 mL50\text{ mL}.
  3. Use the density formula: ρ=55 g50 mL=1.1 g/mL\rho = \frac{55\text{ g}}{50\text{ mL}} = 1.1\text{ g/mL}

Explanation:

The mass of the liquid is isolated by subtracting the weight of the container from the total weight. Density is then found by dividing this mass by the measured volume.

Problem 2:

If a sugar solution has a density of 1.25 g/cm31.25\text{ g/cm}^3, what would be the mass of 200 cm3200\text{ cm}^3 of this solution?

Solution:

Given: Density (ρ\rho) = 1.25 g/cm31.25\text{ g/cm}^3 Volume (VV) = 200 cm3200\text{ cm}^3 Using the formula: m=ρ×Vm = \rho \times V m=1.25×200m = 1.25 \times 200 m=250 gm = 250\text{ g}

Explanation:

By rearranging the density formula ρ=mV\rho = \frac{m}{V}, we can solve for mass by multiplying density and volume.

Determination of density Class 8 Notes & Examples