Answer:
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:
[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].
In this case:
Replacing:
[tex]0.32 M=\frac{number of moles of solute}{1.5 L}[/tex]
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?
[tex]mass=\frac{0.48 moles*84 grams}{1 mole}[/tex]
mass= 40.32 grams
40.32 grams of solute required for the preparation of 1.5L of 0.32M NaHCO₃
