What is the maximum volume of 0.788m cacl2 solution that can be prepared using 85.3g cacl2

Concentration=moles/volume
Volume=moles/concentration
To find the number of moles, you divide the mass of CaCl2 by its molar mass. The molar mass of CaCl2 is: 2x35.45(molar mass Cl) + 40.08(molar mass Ca)
Molar mass= 110.98g/mol
Moles=m/molar mass
Moles=85.3/110.98
Moles=0.7686069562
Then you calculate the volume:
Volume= 0.7686069562/0.778
Volume= 0.9879266789L
=0.988L
Therefor the max volume that can be prepared is 0.998L

Ch4 is nonpolar and insoluble in water. since all of the hydrogen atoms are evenly spread around the carbon atom even through the c h bond is polar all the poles cancel each other out.

6because i no yea yea