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1-liter 10x stock solution of SDS-running buffer is made. The 10x solution has 1%(w/v) concentration of SDS(MW288.8). What is the molarity of the SDS in 1x running buffer? the 10x stock is made at a concentration that is 10 times that of the concentration

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

Start by looking at the 10X solution. You know that this must have a 1% w/v 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 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 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 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 the 10X solution ten-fold.

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

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

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