(A) Calculate the molarity of a solution that contains 0.0345 mol NH4CL In 400 mL of solution. (B) How many moles of HNO3 are present in 35.0 mL of a 2.20 M solution of nitric acid? (C) How many milliliters of 1.50 M KOH solution are needed to supply 0.125 mil of KOH ?

Respuesta :

Answer:

Step-by-step explanation:

(A) We know that Molarity=[tex]\frac{moles solute}{litres solution}[/tex]

Since, moles solute= 0.0345 and litres solution =400mL=0.400l, thus

Molarity=[tex]\frac{3.45{\times}10^-2}{0.400}[/tex]

=[tex]8.62{\times}10^{-2}[/tex]

=[tex]0.0862M[/tex]

(B) Moles solute=Molarity×litres solution

=2.20×0.0350

=0.077 moles

(C) Using the formula,

Litres solution=[tex]\frac{moles of solute}{molarity}[/tex]

=[tex]\frac{0.125}{1.50}[/tex]

=0.0833L=83.33mL