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3 years ago

Name three properties of solids that are different from those of liquids. Explain the differences for each.

1 answer:

4 0

1) Solids have a fixed shape
The particles of solids are held into their positions and are only able to vibrate about fixed points. The strength of the bonding means that the particles have their own shape, which they retain, and do not take the form of the container they are in.

2) Solids may not flow
The molecules of liquids are able to slide past one another due to weaker bonds between particles. This allows the liquid to flow; whereas, this movement is not possible in solids so they cannot flow.

3) Solids may fracture
The fact that solids possess a rigid structure means that they have the ability to fracture. Although some are able to withstand more stress than others, all solids may break. This breakage occurs when the intermolecular bonds are mechanically broken. This is not possible in the case of liquids because such rigid intermolecular bonds do not exist.

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Read 2 more answers

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

Answer: The freezing point of solution is -0.454°C

Explanation:

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

Hence, the freezing point of solution is -0.454°C

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3 years ago