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

Answer:

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:

$\begin{gathered} V__2=\frac{C_1V_1}{C_2} \\ V_2=\frac{500\times350}{100} \\ V_2=\frac{175,000}{100} \\ V_2=1750mL \end{gathered}$

Hence the amount of water needed is 1750mL