# 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

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.

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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