What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g of CaCl2?
Answer (2)
The answer is 0.975 L
Volume = mol/Molarity
We have molarity (0.788 M) and we need mol and volume. Let's first calculate number of moles of CaCl2 in 85.3 g:
Molar mass of CaCl2 is sum of atomic masses of Ca and Cl: Mr(CaCl2) = Ar(Ca) + 2Ar(Cl) = 40 + 2 * 35.45 = 40 + 70.9 = 110.9 g/mol
So, if 110.9 g are in 1 mol, 85.3 g will be in x mol: 110.9 g : 1 mole = 85.3 g : x x = 85.3 g * 1 mole / 110.9 x = 0.769 moles
Now, calculate the volume: V = 0.769/0.788 V = 0.975 L
The maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g of CaCl2 is approximately 0.975 liters. This is calculated by first determining the moles of CaCl2 in the given mass and then applying the molarity formula to find the volume. These calculations require knowledge of molar mass and the basic relationships between mass, moles, and volume.
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