Question

100ml of a solution, prepared by dissolving 1.755g sodium chloride in water, is mixer with a 100ml solution by dissolving 2.775g calcium chloride in water. The molarity of chloride ion in the final solution is
1)0.275 M
2)0.3 M
3)0.4 M
4)0.55M​

Answer 1

NaCl

mass= 1.755 g

molar mass = 58.5 g

moles = 1.755 /58.5

moles of NaCl =0.03

moles of Cl - ions in NaCl = 0.03 ×1 = 0.03

( n factor is 1 as 1 mole of NaCl gives 1 mole of Cl )

BaCl 2

mass= 2.755 g

molar mass = 208.23

moles = 2.755 /208.23

moles of BaCl2 =0.013

moles of Cl - ions in BaCl2 = 0.013 ×2 = 0.026

( n factor is 2 as 1 mole of BaCl2 gives 2 mole of Cl )

total no of moles of Cl -

= 0.03+ 0.026

= 0.056

total volume = 200 ml = 0.2 L

Molarity

= moles / vol of soln

= 0.056/ 0.2

= 0.28 M

Answer 2

Answer:

The molarity of chloride ion in the final solution is 0.28 M.

Explanation:

NaCl

mass= 1.755 g

Molar mass: 58.5 g

moles = 1.755 /58.5

NaCl moles equal 0.03

NaCl contains 0.03 moles of Cl- ions for 1 mole of NaCl.

(n factor is 1, since 1 mole of NaCl produces 1 mole of Cl)

BaCl 2

mass= 2.755 g

Molecular weight: 208.23

moles = 2.755 /208.23

equals 0.013 moles of BaCl2.

0.013 x 2 = 0.026 is the number of moles of Cl-ions in BaCl2.

(n factor = 2 because 1 mole of BaCl2 produces 2 moles of Cl.)moles of Cl total in number

= 0.03+ 0.026

= 0.056

0.2 L (200 ml) is the total volume.

Molarity

= Mole Volume / Mole Volume

= 0.056/ 0.2

= 0.28 M

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