1. Calculate the number of grams of solute required for the preparation of 1.5L of 0.32M NaHCO3 (MW=84) V1 X C1 = V2 X C2

40.32 grams of solute required for the preparation of 1.5L of 0.32M NaHCO₃

Explanation:

Molar concentration or molarity is a measure of the concentration of a solute in a solution. Molarity is defined as the number of moles of solute present in one liter of solution.

Molarity is calculated by the expression:

Molarity is expressed in units .

In this case:

Replacing:

Solving:

number of moles of solute= 0.32 M* 1.5 L

number of moles of solute= 0.48 moles

Being the molar weight of NaHCO₃ equal to 84 g / mole, the following rule of three can be applied: if there are 84 grams in 1 mole, how much mass is there in 0.48 moles?

mass= 40.32 grams

40.32 grams of solute required for the preparation of 1.5L of 0.32M NaHCO₃

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explanation:

2b2t

gasoline is a mixture. it contains several different kinds of compounds that can be present in a variety of ratios.