Question

(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 ?

Answer 1

Explanation:

(A) We know that Molarity=
`(moles solute)/(litres solution)`

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

Molarity=
`\frac{3.45{*}10^-2}{0.400}`

=
`8.62{*}10^(-2)`

=
`0.0862M`

(B) Moles solute=Molarity×litres solution

=2.20×0.0350

=0.077 moles

(C) Using the formula,

Litres solution=
`(moles of solute)/(molarity)`

=
`(0.125)/(1.50)`

=0.0833L=83.33mL