A student is running an experiment in which 60.3 grams of LiCl is needed, but the only jar of reagent in the lab is labelled lithium chloride monohydrate. How many grams of the hydrate must the student weigh out in order to get the desired amount of the anhydrous compound?

Mass of the hydrated salt required is 85.8 g

Explanation:

Molar mass of anhydrous lithium chloride = 42.5 g/mol

Molar mass of lithium chloride monohydrate = 60.5 g/mol

Mass of water molecule in the hydrated salt = 18 g/mol

Ratio of lithium chloride to water molecule in the hydrated salt = 42.5 : 18

Mass of water molecule in the hydrated salt that will contain 60.3 g of lithium chloride = 60.3/42.5 × 18 g = 25.5 g

Therefore mass of hydrated salt required = (60.3 + 25.5) g = 85.8 g

The student will have to weigh out 85.8 g of the hydrated salt, and then heat the salt in an evaporating dish until it decomposes to liberates all the water of hydration in order to obtain the anhydrous salt.

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.