Yes, it does. The freezing point of the salt solution is lower than the freezing point of pure water since there will be more particles, which are water molecules plus salt molecules, present in the salt solution. The salt molecules will slow down the amount of crystals formed resulting to a lower freezing point.
The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is : - 0.69°C
<h3>Determine the freezing point of the solution </h3>
First step : Calculate the molality of NaCl
molality = ( 2.5 grams / 58.44 g/mol ) / ( 230 * 10⁻³ kg/ml )
= 0.186 mol/kg
Next step : Calculate freezing point depression temperature
T = 2 * 0.186 * kf
where : kf = 1.86°c.kg/mole
Hence; T = 2 * 0.186 * 1.86 = 0.69°C
Freezing point of the solution
Freezing temperature of solvent - freezing point depression temperature
0°C - 0.69°C = - 0.69°C
Hence the Freezing temperature of the solution is - 0.69°C
Learn more about The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is : - 0.69°C
NaCl molar mass 58.4 g/mol
moles of NaCl = .35 g * 1 mol/58.4 g = 0.005993 mol
volume of solution = 150 mL = 0.15 L
molarity = mol/vol(in L)
[NaCl] = 0.005933mol/0.15L = 0.03995 M = 0.040 M
Answer:
filtration
Explanation:
1.put the mixture in a beaker
2.add water to the beaker and stir to dissolve salt
3.filter using a filter paper to have sand as the residue
4.Evaporate the water to remain with salt crystals
