A pipet delivers 9.98 g of water at 20°C. What volume does the pipe deliver?

The idea here is that water's density is temperature dependent, meaning that it varies depending on the temperature of the water sample.

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This is why the problem provides you with the temperature of the water. Water has a maximum density of approximately #"1.00 g mL"^(-1)# at #4^@"C"# #-># you can read more about that in this Socratic answer.

So, as temperature increases, the density of water decreases. This means that you will get less than #"1.00 g"# per milliliter of water as you increase temperature.

I assume that the problem provided you with a table of water densities a various temperatures, but since you don't have it listed here, I'll use this resource to find it

http://www.simetric.co.uk/si_water.htm

So, at #20^@"C"# water is said to have a density of #"0.998203 g mL"^(-1)#.

You can use the density of water as a conversion factor to determine what volume would #"9.98 g"# of water occupy at this temperature

#9.98 color(red)(cancel(color(black)("g"))) * overbrace("1 mL"/(0.998203color(red)(cancel(color(black)("g")))))^(color(brown)("density at"color(white)(a)20^@"C")) = "9.99797 mL"#

Round to three sig figs, the number of sig figs you have for the mass of the sample, the answer will be

#V_"water" = color(green)(|bar(ul("10.0 mL"))|) -># at a temperature of #20^@"C"#