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?
Answers
Answer:
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.
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Explanation:
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(3) evaporation of the solvent
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Answers
Answer:
4) titration
Explanation:
Titration is a standard process used in a laboratory to determine the concentration of an unknown analyte. A titrant of known concentration is gradually added to a known volume of the analyte in the presence of a suitable indicator. The end of the titration is marked by a color change of the analyte.
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M2 and V2 are the molarity and volume of HBr. Here, V2 is known whereas M2 needs to be determined.
Based on the reaction stoichiometry:
moles of NaOH = moles of HBr
Answers
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The powder heats up.
The powder breaks apart into smaller particles.
Answers
Answer:
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Explanation:
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