Question

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

Answer 1

`"10.0 mL"`

Explanation:

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

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"`