Question #a6a3b

Your answer is a little off because the total mass of the solution includes the mass of the added sugar.

You start with #"100 mL"# of water at a density of #"1 g/mL"#, which means that you have #"100 g"# of water.

When you add the sugar, the mass of the solution will increase

#m_"solution" = m_"water" + m_"sigar"#

#m_"solution" = 100 + 4 = "104 g"#

A solution's percent cocnetration by mass is calculated by dividing the mass of the solute, in your case sugar, by the total mass of the solution, then multiplying the result by 100.

#"% w/w" = m_"solute"/m_"solution" * 100#

In your case, this means

#m_"sugar"/m_"solution" * 100 = "% w/w"#

#(4color(red)(cancel(color(black)("g"))))/(104color(red)(cancel(color(black)("g")))) * 100 = color(green)(3.85%)#

Now, an important thing to look out for is the number of sig figs. As it is written, the information you provide justifies an answer rounded to one sig fig - this is because you only gave one sig fig for the volume of water.

In this case, the answer would indeed by

#"% w/w" = 4#

However, I assume that you are supposed to use three sig figs for the answer, which implies that the volume of water is actually #"100. mL"#.

Anyway, the important thing to remember is that for percent concetrations by mass you need the mass of the solute and the mass of the solution, i.e. solute + solvent.