Question

A solution contains 0.9% of Sodium Chloride. Calculate the number of particles that contribute to the osmotic pressure in 1 Liter of the solution. (Mol wt: 58.5). Assume that the sodium chloride is completely soluble and each mole of NaCl dissociates in two ions (Nat & Cl") at this concentration. 0.99 x 9 m - 1 liter a. Calculate the quantity of sodium chloride present in 1 liter of solution. Toomi x=9grams b. How many molecules are present in 1 liter? c. How many ions (particles) are formed when 1 mole of sodium chloride dissociates? d. How many ions (particles) are present in 0.9% sodium chloride?

A solution contains 0.9% of Sodium Chloride. Calculate the number of particles that contribute to the osmotic pressure in 1 Liter of solution. (Mol wt: 58.5). Assume that the sodium chloride is completely soluble and each mole of NaCl dissociates in two ions (Na +& Cl) at this concentration.

1. Calculate the quantity of sodium chloride present in 1 liter of solution.

2. How many molecules are present in 1 liter?

3. How many ions (particles) are formed when 1 mole of sodium chloride dissociates?

4. How many ions (particles) are present in 0.9% sodium chloride?

Answer 1

The number of particles that contribute to the osmotic pressure in 1 Liter of solution = 1.20 x 10²⁴

0.9 % means 0.9 grams of NaCl present in 100 mL of solution

a)

0.9 g / 100 mL ==> 9 grams in 1000 mL = 1 L

Answer: the quantity of sodium chloride present in 1 liter of solution = 9 g

b)

Avogadro's number N_A = 6.02214129 x 10²³ formula units/mol

moles NaCl in 9g = 9g / 58.5 g/mol = 0.15384615384 mol

molecules present in 1 L of solution = 0.15384615384 mol x 6.02214129 x 10²³ molecules = 9.26 x 10²²

Answer: molecules present in 1 L of solution = 9.26 x 10²²

c)

Each molecule of NaCl dissociates into 2 ions NaCl ==> Na⁺ + Cl^-

1 mole gives 2 ions, so ions formed when 1 mole of sodium chloride dissociates = 2 x 6.02214129 x 10²³

Answer: ions (particles) are formed when 1 mole of sodium chloride dissociates = 1.20 x 10²⁴

d)

moles in 0.9 g = 0.9g / 58.5 g/mol = 0.015384615384 mol

molecules present in 0.9% of solution = 0.015384615384 mol x 6.02214129 x 10²³ molecules = 9.26 x 10²¹

ions formed = 2 x 0.015384615384 mol x 6.02214129 x 10²³ molecules = 1.85 x 10²²

Answer: ions (particles) are present in 0.9% sodium chloride = 1.85 x 10²²

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