
Concept explainers
a.
Interpretation:
Mass in grams of solute needed to prepare 3.50 L of 0.500 M should be determined.
Concept introduction:
The concentration of any solution means the amount of solute present in the specific volume of the solvent. Now the amount of solute present in the solution is expressed by different units like gram, gram-equivalent, gram-mole or mole etc. The amount of solvent to prepare the solution can be expressed in terms of weight or volume. The different unit of expression the concentration of a solution are- percentage strength, normality, molarity, molality, formality, gram per litter, mole fraction, parts per million etc.
- Molarity: In per litter (1000 mL or 1000 cc) volume of any solution the amount of solute present in gram-mole is called the molarity of the solution. It is expressed by ‘M’. Like- If in 1 L of
solution 2 gm-mole of the pure
is present then the strength of the solution will be 2 M. It can be expressed as-
b.
Interpretation:
Mass in grams of solute needed to prepare 65.0 mL of 1.45 M should be determined.
Concept introduction:
The concentration of any solution means the amount of solute present in the specific volume of the solvent. Now the amount of solute present in the solution is expressed by different units like gram, gram-equivalent, gram-mole or mole etc. The amount of solvent to prepare the solution can be expressed in terms of weight or volume. The different unit of expression the concentration of a solution are- percentage strength, normality, molarity, molality, formality, gram per litter, mole fraction, parts per million etc.
- Molarity: In per litter (1000 mL or 1000 cc) volume of any solution the amount of solute present in gram-mole is called the molarity of the solution. It is expressed by ‘M’. Like- If in 1 L of
solution 2 gm-mole of the pure
is present then the strength of the solution will be 2 M. It can be expressed as-

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Chapter 5 Solutions
Chemistry For Changing Times (14th Edition)
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- A student proposes the transformation below in one step of an organic synthesis. There may be one or more reactants missing from the left-hand side, but there are no products missing from the right-hand side. There may also be catalysts, small inorganic reagents, and other important reaction conditions missing from the arrow. • Is the student's transformation possible? If not, check the box under the drawing area. . If the student's transformation is possible, then complete the reaction by adding any missing reactants to the left-hand side, and adding required catalysts, inorganic reagents, or other important reaction conditions above and below the arrow. • You do not need to balance the reaction, but be sure every important organic reactant or product is shown. + T X O O лет-ле HO OH HO OH This transformation can't be done in one step.arrow_forwardDetermine the structures of the missing organic molecules in the following reaction: X+H₂O H* H+ Y OH OH Note: Molecules that share the same letter have the exact same structure. In the drawing area below, draw the skeletal ("line") structures of the missing organic molecules X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. X Sarrow_forwardPredict the major products of this organic reaction. If there aren't any products, because nothing will happen, check the box under the drawing area instead. No reaction. HO. O :☐ + G Na O.H Click and drag to start drawing a structure. XS xs H₂Oarrow_forward
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