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Nesterboy [21]
3 years ago
15

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

Chemistry
1 answer:
densk [106]3 years ago
8 0

Answer:

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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8 0
2 years ago
500 mL of a solution contains 1000 mg of CaCl2. Molecular weight of CaCl2 is 110 g/mol. Specific gravity of the solution is 0. C
dangina [55]

Answer:

a) 0,2% w/v

b) r=500

c) 0,0182 M

d) 0,0145 m

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Explanation:

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%w/v= \frac{1,000 g CaCl2}{500mL}×100 = 0,2%w/v

b) Ratio strength is a way to express concentration.  For w/v is in 1g of solute <em>r</em> mililiters of solution have. Thus, r = 500 because we have in the first 1 g of CaCl₂ in 500 mL of solution.

c) Molarity is moles of solute per liter of solution, thus:

1,000 g of CaCl₂ × \frac{1mol}{110g} = 9,09×10⁻³ moles of CaCl₂

500 mL of solution  × \frac{1L}{1000mL} = 0,500 L of solution

M = \frac{9,09x10^{-3} moles }{0,500 L} = 0,0182 M

d) Molality is moles of solute per kg of solution.

Specific gravity is the ratio between density of the solution and density of a reference substance (Usually water). With a specific gravity of 0,8:

kg of solution = 0,500 L of solution × \frac{0,8 kg}{1L} =<em> </em><em>0,625 kg of solution</em>

m = \frac{9,09x10^{-3}moles }{0,625 kg} = 0,0145 m

e)  In a salt, equivalents are the number of moles ables to replace one mole of charge. In CaCl₂ is ¹/₂ because with  ¹/₂ moles of CaCl₂ it is possible to replace 1 mole of charges. Thus, in 1,5 L there are:

1,5 L ×\frac{0,0182 CaCl2 moles}{1L} × \frac{1equivalent}{2 moles} = 0,0137 equivalents

I hope it helps!

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3 years ago
Is light an external or internal stimulus for plant?explain how plants react to light
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3 years ago
In the following equation:
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Answer:

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A simple way to determine which reagent is the limiting reactant is to convert all given data to moles then divide by the respective coefficients of the balanced equation. The smaller value will be the limiting reactant.

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\\ \sf\longmapsto No\;of\:moles=3mol

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