Question

A solution is prepared by adding 1. 60 g of solid nacl to 50. 0 ml of 0. 100 m cacl2. What is the molarity of chloride ion in the final solution? assume that the volume of the final solution is 50. 0 ml.

Answer 1

The number of moles of a solute per liter of a solution exists comprehended as molarity. The molarity of chloride ion in the final solution 0.747 M.

What is meant by molarity?

The amount of a substance in a specific volume of solution is known as its molarity (M). The number of moles of a solute per liter of a solution is known as molarity.

CaCl2 has a molarity of 0.100 M.

CaCl2 volume = 50.0 mL = 0.05 L

NaCl has a mass of 1.60 g.

The solution's final volume exists 50.0 mL, or 0.05 L.

Number of moles of CaCl2 = Molarity × Volume 0.100 × 0.05 L

= 0.05 mol

One mole of Ca2+ and two moles of Cl-ions are produced upon dissociation of CaCl2.

Moles of chloride ions = 2 moles of calcium chloride

= 2 × 0.05 moles = 0.01 moles.

On calculating the moles of NaCl, we get

n = mass/molar mass

= 1.60g/ 58.4 g/mol

= 0.0274 mol

One mole of N + and one mole of Cl + ions are produced upon dissociation of NaCl.

As a result, moles of chloride ion are equal to moles of NaCl and equal to 0.0274 moles.

The solution has a total of 0.01 mol + 0.0274 mol chloride ions, or 0.0373 mol.

Given this, the final solution's volume is 0.05 L.

Molarity of chloride ions = moles/volume

= 0.0373 mol/ 0.05 L

= 0.747 M

The final solution contains 0.747 M of chloride ions as a result.

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Answer 2

The molarity of chloride ion in the 50.0 ml final solution prepared by adding 1. 60 g of solid NaCl to 50. 0 ml of 0. 100 m CaCl₂ is 0.747 M.

What is NaCl?

Although sea salt also contains other chemical salts, sodium chloride (NaCl), generally known as salt, is an ionic compound with the chemical formula NaCl, denoting a 1:1 ratio of sodium and chloride ions. The molar weights of 39.34 g Na and 60.66 g Cl in 100 g of NaCl are 22.99 and 35.45 g/mol, respectively.

As given in the question,

CaCl₂ has a molarity of 0.100 M.

CaCl₂ volume = 50.0 mL = 0.05 L

NaCl has a mass of 1.60 g.

The solution's final volume is 50.0 mL, or 0.05 L.

Lets calculate the number of moles of CaCl₂,

$N = M \times V$

Where, N = no. of moles, V = volume, M = molarity

N = 0.100 x 0.05 L

N = 0.05 mol

One mole of Ca2+ and two moles of Cl- ions are produced upon dissociation of CaCl₂. As a result,

moles of chloride ions = 2 moles of calcium chloride

Moles of chloride ions = 2 times 0.05 moles = 0.01 mol.

Lets calculate the number of moles of NaCl

$N = M \times V$

N = 0.0274 mol

One mole of N + and one mole of Cl + ions are produced upon dissociation of NaCl. As a result,

Moles of chloride ion are equal to moles of NaCl

Moles of Chloride equal to 0.0274 mol.

Total moles of chloride = 0.01 mol + 0.0274

Total moles of chloride = 0.0373 mol

Total volume given in question is 0.05 L.

So, Molarity will be:

$M = \frac{N}{V}$

M = 0.0372/0.05 mol/l

M = = 0.747 M

Therefore, The final solution contains 0.747 M of chloride ions as a result.

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