Im doing a lab and I\'ve been working on this problem for a long time and I can\
ID: 801339 • Letter: I
Question
Im doing a lab and I've been working on this problem for a long time and I can't figure out what I'm doing wrong.
(I keep getting 99% empty space and I know that it's wrong)
Given:
Assume the volume of water is the additive volume of its atoms
H2O amu: 18.0153
Hydrogen atomic radious: 37pm
Oxygen atomic radious: 66pm
Equation: % Empty Space = ( (Va -Vs) / Va ) x 100
(Va= Volume of arragement, Vs=Volume of sphere)
Here's my info:
Mass of 50mL beaker----------------- 73.128 g
Volume of water------------------------- 25 mL
Mass of 50mL beaker and water-- 97.815 g
Mass of Water--------------------------- 24.517 g
Density of water------------------------- .98 mL/g
Volume of 1 water molecule--------- (4/3?)(140 x 10^-12)^3 = 1.15 x 10^-29
Total number of water molecules in 1mL H20 ----- (24.517/18.0153)(6.022 x 10^23) = 8.2 x 10^23
Total volume of water molecules (V spheres)------- (.98 x (1.15 x 10^23)) = 9.425 x 10^-6
Water percent empty space------------------------------- ?
Explanation / Answer
Volume of water------------------------- 25 mL
Mass of 50mL beaker and water-- 97.815 g
Mass of Water--------------------------- 24.517 g
Density of water------------------------- .98 mL/g
Volume of 1 water molecule--------- (4/3?)(140 x 10^-12)^3 = 1.15 x 10^-29
Mass of 1 ml water=1/0.98=1.020408g
Total number of water molecules in 1mL H20=N= ----- (1.020408/18.0153)(6.022 x 10^23) = 3.410933 x 10^22
Total volume of sphere=(1.15*10^-29)*(3.410933*10^22)=3.9226*10^(-7)m^3=0.392ml
Void fraction=(1-0.392)/1=0.608=60.8%