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A reagent bottle contains a stock solution of 0.90% by mass of sodium chloride, NaCl.
The density of the solution is 1.00 g cm ?. Calculate
(i) the mole fraction of NaCl
(ii) the molality of NaCl solution
(111) the volume of the stock solution required to prepare 100 mL of 0.01 M NaCl
solution.​

Respuesta :

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Answer:

I. 2.79x10⁻³

II. 0.155m

III. 6.49mL of the solution you must add from the stock solution

Explanation:

A solution of 0.90% by mass NaCl means that, in 100g of solution, you have 0.90g of NaCl. In other words, there are 0.90g of NaCl + 99.1g of water.

I. Mole fraction of NaCl is the ratio between moles of NaCl and total moles of solution.

Moles NaCl:

0.90g * (1mol / 58.44g) = 0.0154 moles NaCl

Moles water:

99.1g water * (1mol / 18.01g) = 5.50 moles water.

Total moles = 5.50 +0.0154 moles = 5.518 moles

Mole fraction NaCl: 0.0154 / 5.518 moles =

2.79x10⁻³

II. Molality is the ratio between moles of solute (NaCl) and kg of solvent.

There are 99.1g of water = 0.0991kg.

Molality is: 0.0154 moles / 0.0991kg

0.155m

II. 100mL of a 0.01M are:

100mL = 0.100L * (0.01 mol / L) = 1x10⁻³ moles of NaCl

In 100g = 100mL (Density of 1g/mL) there are 0.0154 moles. To have 1x10⁻³ moles you need:

1x10⁻³ moles of NaCl * (100mL / 0.0154 moles) =

6.49mL of the solution you must add from the stock solution

I. Mole fraction of NaCl is 2.79x10⁻³

II. Molarity of NaCl solution is 0.155m

III. 6.49mL of the stock solution required to prepare 100 mL of 0.01 M NaCl  solution.​

What does % by mass mean?

A solution of 0.90% by mass NaCl means that, in 100g of solution, you have 0.90g of NaCl.

Or

There is 0.90g of NaCl + 99.1g of water.

We need to find:

I. Mole fraction of NaCl:

Mole fraction can is the ratio of moles of Solute  to the total number of moles of the solutes and the solvent.

Moles of  NaCl(solute) = 0.90g * (1mol / 58.44g) = 0.0154 moles NaCl

Moles of  water(solvent) = 99.1g water * (1mol / 18.01g) = 5.50 moles water.

Total moles = 5.50 +0.0154 moles = 5.518 moles

Mole fraction of NaCl =  [tex]\frac{0.0154}{ 5.518} =2.79 * 10^{-3}[/tex]

II. Molality is a measure of the number of moles of solute(NaCl) in a solution corresponding to 1 kg or 1000 g of solvent

Since, 99.1g of water = 0.0991kg.

Molality of NaCl = [tex]\frac{0.0154 moles}{0.0991 kg}=0.155m[/tex]

III. 100mL of a 0.01M is:

100mL = 0.100L * (0.01 mol / L) = 1x10⁻³ moles of NaCl

In 100g = 100mL (Density of 1g/mL) there are 0.0154 moles. To have 1x10⁻³ moles you need:

1x10⁻³ moles of NaCl * [tex]\frac{100 mL}{0.0154 moles}[/tex] = 6.49mL of the solution you must add from the stock solution

Learn more:

https://brainly.com/question/15063496

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