Answered

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what is the maximum volume of 0.788M CaCl2 solution that can be prepared using 85.3g CaCl2

Sagot :

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

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