Answer
205 mL is the volume that this mass of water would have at $80^{o}$ C.
1L water at $20^{o}C$ has more mass than 1L water at $80^{o}C$:
$m_{at20^{o}c}$$\gt$$m_{at80^{o}c}$
Therefore there are more water molecules in 1L water at $20^{o}C$ than in 1L water at $80^{o}C$.
Work Step by Step
Strategy: We are given density of water for 2 temperatures $20^{o}$ C and $80^{o}$ C. As we are given a volume 200mL of water at $20^{o}$ C, we can find the mass .
This mass at $80^{o}$ C would have another volume. we can find it by formula:
V = $\frac{m}{d }$.
Note however, that the densities at $20^{o}$ C and $80^{o}$ C are given in $\frac{kg}{m3}$, so we have to convert these densities to $\frac{g}{mL}$, in order to find the volume of water at $80^{o}$ C in mL.
Step 1: Covert densities from $\frac{kg}{m3}$ to $\frac{kg}{mL3}$.
Density at $20^{o}$ C : 998 $\frac{kg}{m3}$ =998 $\frac{10^{3}g}{10^{6}mL}$
998 $\frac{kg}{m3}$ = 998 x $10^{-3}$ $\frac{g}{mL}$ = 0.998$\frac{g}{mL}$
Density at $80^{o}$ C: 972 $\frac{kg}{m3}$ = 972 $\frac{10^{3}g}{10^{6}mL}$
972 $\frac{kg}{m3}$ = 972 x $10^{-3}$$\frac{g}{mL}$ = 0.972$\frac{g}{mL}$
Step2: find the mass of water when we know the density at $20^{o}$ C and the volume : m = d x V
$m_{water}$ = 0.998 $\frac{g}{mL}$ x 200mL
$m_{water}$ = 199.6 g
Now find the volume that this mass of water would have at $80^{o}$ C
V = $\frac{m}{d }$
V = $\frac{199.6 g}{0.972\frac{g}{mL} }$
V = 205.35mL 0r V = 205 mL.
1 L water at $20^{o}$ C would contain more water molecules than 1L water at $80^{o}$ C.
we know m = d x V. Volume is 1L or $10^{3}$mL:
$m_{at20^{o}c}$ = 0.998$\frac{g}{mL}$ x $10^{3}$mL
$m_{at20^{o}c}$= 998 g
$m_{at80^{o}c}$ = 0.972$\frac{g}{mL}$ x $10^{3}$mL
$m_{at20^{o}c}$= 972 g
So 1L water at $20^{o}C$ has more mass than 1L water at $80^{o}C$:
$m_{at20^{o}c}$$\gt$$m_{at80^{o}c}$
Therefore there are more water molecules in 1L water at $20^{o}C$ than in 1L water at $80^{o}C$.
