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In Chemistry / High School | 2014-11-24

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

Asked by Anonymous

Answer (2)

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

Answered by baseballbeast | 2024-06-10

The maximum volume of a 0.788 M CaCl₂ solution that can be prepared from 85.3 g of CaCl₂ is approximately 0.966 liters (966 mL). This is calculated by determining the number of moles in the given mass and then using the molarity formula to find the volume. The molar mass of CaCl₂ is about 111.98 g/mol, which is necessary for these calculations.
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Answered by baseballbeast | 2024-12-16

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