Question

21. If you have 350 milliliters (mL) of a 500-milligramper liter (mg/L) solution of alum (aluminum sulfate),how much water would you have to add to bring theconcentration down to 100 mg/L? Assume that youare adding water that does not contain any alum (C2=0).1. 70 mL2. 1,400 mL3. 143 mL4. 1,750 mL

Answer 1

1750mL

Explanations:

In order to get the required volume, we will use the dilution formula expressed as:

`C_1V_1=C_2V_2`

where:

C₁ and C₂ are the initial and final concentration respectively

V₁ and V₂ are the initial and final volume respectively

Given the following parameters:

C₁ = 500mg/L

C₂ = 100mg/L

V₁= 350mL

Required

V₂

Substitute the given parameters into the formula to have:

`\begin{gathered} V__2=(C_1V_1)/(C_2) \\ V_2=(500*350)/(100) \\ V_2=(175,000)/(100) \\ V_2=1750mL \end{gathered}`

Hence the amount of water needed is 1750mL