What is the concentration of chloride in a solution made with 0.808 grams of CaCl2 and 250.0 ml of water.?

0.0584 M

Explanation:

From the question given above, the following data were obtained:

Mass of CaCl₂ = 0.808 g

Volume of water = 250 mL

Concentration of chloride =?

Next, we shall determine the number of mole in 0.808 g of CaCl₂. This can be obtained as follow:

Mass of CaCl₂ = 0.808 g

Molar mass of CaCl₂ = 40 + (35.5 × 2)

= 40 + 71

= 111 g/mol

Mole of CaCl₂ =?

Mole = mass / Molar mass

Mole of CaCl₂ = 0.808 / 111

Mole of CaCl₂ = 0.0073 mole

Next, we shall convert 250 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

250 mL = 250 mL × 1 L / 1000 mL

250 mL = 0.25 L

Next, we shall determine the molarity of CaCl₂. This can be obtained as follow:

Mole of CaCl₂ = 0.0073 mole

Volume of water = 0.25 L

Molarity of CaCl₂ =?

Molarity = mole /Volume

Molarity of CaCl₂ = 0.0073 / 0.25

Molarity of CaCl₂ = 0.0292 M

Finally, we shall determine the concentration of the chloride as illustrated below:

CaCl₂ <=> Ca²⁺ + 2Cl¯

From the equation above,

1 mole of CaCl₂ produced 2 mole of Cl¯.

Therefore, 0.0292 M CaCl₂ will produce = 0.0292 × 2 = 0.0584 M Cl¯.

Thus, the concentration of the chloride ion (Cl¯) in the solution is 0.0584 M

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explanation:

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"