It is always true that the rate at which a solute dissolves can be increased by grinding. The smaller the solute the easier it will dissolve in the solvent, while other facts play into the rate at which a solute dissolves in a solvent, a major part of this is also how small the solute is. You can think of how rock salt is harder to dissolve in water compared to finely ground salt.
It is sometimes true that as the temperature of a solvent decreases, the solubility of a solute increase. The reason for this is that for liquids and solids as temperature increases the solubility increases but for gasses, as the temperature increases the solubility decreases.
It is always true that stirring a solute when adding it to a solvent should increase the rate of its dissolving. however, this will not increase the amount that is able to be dissolved in the solution.
It is never true that Henry's law states that the solubility of a gas in a liquid is a function of temperature. Henry's law is a gas law that was determined by William Henry in 1803. The law dictates that when in constant temperature the amount of gas that dissolves in a given volume of a liquid is proportional directly to the partial pressure of the gas at equilibrium with the desired liquid. In simpler terms, the solubility of the gas in a certain liquid is proportional to the partial pressure of the gas above the liquid.
It is always true that two liquids that dissolve in each other are miscible. Miscibility is described as the property of liquids and other substances to mix in all proportions and forming homogeneous solutions.
Answer:
0.001 mole of NaF.
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 100 mL
Molarity = 0.01 M
Mole of NaF =?
Next, we shall convert 100 mL to litre (L). This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100 mL × 1 L / 1000 ml
100 mL = 0.1 L
Thus, 100 mL is equivalent to 0.1 L.
Finally, we shall determine the number of mole of NaF present in the solution. This can be obtained as follow:
Volume of solution = 0.1 L
Molarity = 0.01 M
Mole of NaF =?
Molarity =mole /Volume
0.01 = mole of NaF / 0.1
Cross multiply
Mole of NaF = 0.01 × 0.1
Mole of NaF = 0.001 mole.
Thus, 0.001 mole of NaF is present in 100 mL of the solution.
