Question #24972

The molarity of the 1X Running Buffer will be 0.0035 M.

Start by looking at the 10X solution. You know that this buffer must have a 1% w/v percent concentration of sodium dodecyl sulfate (SDS), and that you have 1-L of solution.

A 1% w/v solution will contain 1 g of SDS in 100 mL of solution, which means that, in order to keep the percent concentration by mass unchanged, a larger volume will require more SDS present.

`"%w/v" = "grams of SDS"/"100 mL" * 100`

`"%w/v" = x/("1000 mL") * 100 => x = ("%w/v" * "1000 mL")/(100)`

`x = (1 * 1000)/100 = "10 g SDS"`

To determine the molarity of the 10X solution, use the molar mass given to calculate how many moles of SDS are present

`10cancel("g SDS") * "1 mole SDS"/(288.8cancel("g SDS")) = "0.03463 moles SDS"`

Since you have 1 L of solution, you'll get

`C = n/V = "0.03463 moles"/"1 L" = "0.03463 M"`

Because the concentration of a 10X solution is 10 times higher than the concentration of the 1X solution, the molarity of the 1X solution will be

`C_"1X" = C_"10X"/10 = "0.03463 M"/10 = "0.003463 M"`

In other words, the 1X solution is obtained by diluting the 10X solution ten-fold.

I will leave the answer rounded to two sig figs, although your data indicates that it should be rounded to one sig fig

`C_"1X" = color(green)("0.0035 M")`