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A solution contains 2.38 g of magnesium
chloride, MgCl2, in 500 cm3 of solution.
(a) Calculate the number of moles of
magnesium chloride in 500 cm3 of the
solution.
(b) What is the concentration of
(i) magnesium chloride in mol/dm3;
(ii) chloride ions in g/dm3?

Respuesta :

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Answer:

a. 0.0250 moles MgCl2

b. i. 0.0500mol MgCl2 / dm3

ii. 3.54g Cl-/dm3

Explanation:

a. In 500cm3, the mass of MgCl2 (Molar mass: 95.211g/mol) is 2.38g. The moles are:

2.38g * (1mol / 95.211g) = 0.0250 moles MgCl2

b. The dm3 are (1000cm3 = 1dm3):

500cm3 * (1dm3 / 1000cm3) = 0.500dm3

Concentration is:

0.0250 moles MgCl2 / 0.500dm3

i. 0.0500mol MgCl2 / dm3

ii. Moles Cl-:

0.0250 moles MgCl2 * (2mol Cl- / 1mol MgCl2) = 0.0500 moles Cl-

Mass Cl- -Molar mass: 35.45g/mol-:

0.0500 moles Cl- * (35.45g / mol) = 1.77g of chloride ions

1.77g Cl- / 0.500dm3:

3.54g Cl-/dm3

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