A solution was prepared by dissolving 100 mg of protein X in 100 ml of water. Molecular weight of protein X is 15,000 Da; Avogadro’s number = 6.022x 1023. Calculate the molarity(M) of the resulting solution?

The trick here is to realize that protein #"X"# has a molar mass of #"15,000 g mol"^(-1)#.

First of all, I should mention that the problem is actually giving you the molecular mass of the protein, not its molecular weight.

The term molecular weight is currently used to denote relative molecular mass, which expresses the mass of a given molecule relative to #1/12"th"# of the mass of single carbon-12 atom. This implies that molecular weight is unitless.

On the other hand, molecular mass is expressed in unified atomic mass units, #"u"#, or daltons, #"Da"#.

More specifically, you have

#color(blue)(ul(color(black)("1 u " = " 1 Da")))#

Now, the thing to remember about the unified atomic mass unit is that it's equivalent to

#color(blue)(ul(color(black)("1 u " = " 1 g mol"^(-1))))#

In this case, a single protein #"X"# has a molecular mass of #"15,000 Da"#, or #"15,000 u"#, which means that it has a molar mass of

#"15,000" color(red)(cancel(color(black)("u"))) * ("1 g mol"^(-1))/(1color(red)(cancel(color(black)("u")))) = "15,000 g mol"^(-1)#

This means that one mole of protein #"X"# has a mass of #"15,000 g"#

Now, use the molar mass of protein #"X"# to calculate how many moles you have in your sample

#100 color(red)(cancel(color(black)("mg"))) * (1color(red)(cancel(color(black)("g"))))/(10^3color(red)(cancel(color(black)("mg")))) * "1 mole X"/("15,000"color(red)(cancel(color(black)("mg")))) = 6.667 * 10^(-6)"moles X"#

Since molarity is defined as the number of moles of solute present in one liter of solution, the molarity of your solution will be

#c = (6.667 * 10^(-6)"moles X")/(100 * 10^(-3)"L") = 6.667 * 10^(-5)"mol L"^(-1)#

I'm assuming that you need the molarity in micromoles per liter, #mu"mol L"^(-1)#

#6.667 * 10^(-5)color(red)(cancel(color(black)("mol"))) "L"^(-1) * (10^6mu"mol")/(1color(red)(cancel(color(black)("mol")))) = color(darkgreen)(ul(color(black)(67color(white)(.)mu"mol L"^(-1))))#

I'll leave the answer rounded to two sig figs.