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IceJOKER [234]
3 years ago
7

A solution is prepared by dissolving 33.0 milligrams of sodium chloride in 1000. L of water. Assume a final volume of 1000. lite

rs. Calculate the following values listed below.
a. Molarity of NaCl
b. Molarity of sodium ions
c. Molarity of chloride ions
d. Osmolarity of the solution
e. Mass percent of NaCl
f. Parts per million of sodium chloride
g. Parts per billion of sodium chloride
h. Look at your answers to parts e, f, and g.
Which one of these values is the most convenient or easiest to say and use/understand when discussing concentration.
Chemistry
1 answer:
zheka24 [161]3 years ago
5 0

Answer:

a. Molarity of NaCl solution = 5.64 * 10⁻⁷ mol/L

b. molarity of Na⁺ = 5.64 * 10⁻⁷ mol/L

c. molarity of Cl⁻ = 5.64 * 10⁻⁷ mol/L

d. Osmolarity = 1.128 osmol

e. mass percent of NaCl = 3.30 * 10⁻⁶ %

f. parts per million NaCl = 0.033 ppm NaCl

g. parts per billion of NaCl = 33 ppb of NaCl

h. From the values obtained from e, f and g, the most convenient to use and understand is parts per billion as it has less of a fractional part to deal with especially since the solute concentration is very small.

Explanation:

Molarity of a solution = number of moles of solute (moles)/volume of solution (L)

where number of moles of solute = mass of solute (g)/molar mass of solute (g/mol)

a. Molarity of NaCl:

molar mass of NaCl = 58.5 g/mol, mass of NaCl = 33.0/1000) g = 0.033g

number of moles of NaCl = 0.033/58.5 = 0.000564 moles

Molarity of NaCl solution = 0.000564/1000 = 5.64 * 10⁻⁷ mol/L

b. Equation for the dissociation of NaCl in solution: NaCl ----> Na⁺ + Cl⁻

From the above equation I mole of NaCl dissociates to give 1 mole of Na⁺ ions,

Therefore molarity of Na⁺ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L

c. From the above equation I mole of NaCl dissociates to give 1 mole of Cl⁻ ions,

therefore molarity of Cl⁻ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L

d. From the above equation, dissociation of NaCl in water produces 1 mol Na⁺ and 1 mole Cl⁻.

Total number of particles produced = 2

Osmolarity of solution = number of particles * molarity of siolution

Osmolarity = 2 * 5.64 * 10⁻⁷ mol/L = 1.128 osmol

e. mass of percent of NaCl = {mass of NaCl (g)/ mass of solution (g)} * 100

density of water = 1 Kg/L

mass of water = 1 Kg/L * 1000 L = 1000 kg

1Kg = 1000 g

Therefore mass of solution in g = 1000 * 1000 = 1 * 10⁶ g

mass percent of NaCl = (0.033/1 * 10⁶) * 100 = 3.30 * 10⁻⁶ %

f. Parts per million of NaCl:

parts per million = 1 mg of solute/L of solution

One thousandth of a gram is one milligram and 1000 ml is one liter, so that 1 ppm = 1 mg per liter = mg/Liter.

Since the density of water is 1kg/L = 1,000,000 mg/L

1mg/L = 1mg/1,000,000mg or one part in one million.

parts per million NaCl = 33.0/1000 L = 0.033 ppm NaCl

g. Parts per billion = 1 µg/L of solution

1 g = 1000 µg

therefore, 33.0 mg = 33.0 * 1000 µg = 3.30 * 10⁴ µg

parts per billion of NaCl = 3.30 * 10⁴ µg/1000 L = 33 ppb of NaCl

h. From the values obtained from e, f and g, the most convenient to use and understand is parts per billion as it has less of a fractional part to deal with especially since the solute concentration is very small.

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I believe it is C, as helium is one of the lightest noble gases making the particles move faster.
5 0
3 years ago
Calculate the mass of beryllium (Be) needed to completely react with 18.9 g nitrogen gas (N2) to produce Bez N2, which is the on
choli [55]

Answer:

C = 18.29 g  

Explanation:

Given data:

Mass of beryllium needed = ?

Mass of nitrogen = 18.9 g

Solution:

Chemical equation:

3Be + N₂    →    Be₃N₂

now we will calculate the number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 18.9 g/ 28 g/mol

Number of moles = 0.675 mol

Now we will compare the moles of nitrogen and Be from balance chemical equation.

                     N₂        :       Be        

                       1          :       3

                  0.675       :      3/1×0.675 = 2.03 mol

Number of moles of Be needed are 2.03 mol.

Mass of Beryllium:

Mass = number of moles × molar mass

Mass = 2.03 mol ×   9.01 g/mol

Mass = 18.29 g  

4 0
2 years ago
Hydrocyanic acid, HCN, is a weak acid. (a) Write the chemical equation for the dissociation of HCN in water. (b) Identify the Br
12345 [234]

Answer: a) HCN(aq.)+H_2O\rightleftharoons H_3O^+(aq.)+CN^-(aq.)

b) HCN : acid CN^- :conjugate base.

And, H_2O : base H_3O^+: conjugate acid.

c) HCN(aq.)+NaOH(aq)\rightleftharoons NaCN(aq.)+H_2O(l)

d) NaCN(aq)\rightarrow Na^+(aq.)+CN^-(aq)

e) NaCN(aq)+HCl(aq)\rightarrow NaCl(aq.)+HCN(aq)

Explanation:

a) Weak acid is defined as the acid which does not completely dissociates when dissolved in water. They have high pH. These releases H^+ ions in their aqueous states.

The equation for the dissociation of HCN acid is given by:

HCN(aq.)+H_2O\rightleftharoons H_3O^+(aq.)+CN^-(aq.)

b) According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

For the given chemical equation:

HCN is loosing a proton, thus it is considered as an acid and after losing a proton, it forms CN^- which is a conjugate base.

And, H_2O is gaining a proton, thus it is considered as a base and after gaining a proton, it forms H_3O^+ which is a conjugate acid.

c) Neutralization reaction is a reaction in which an acid reacts with base to produce salt and water.

HCN(aq.)+NaOH(aq)\rightarrow NaCN(aq.)+H_2O(l)

d) The chemical equation for dissociation of NaCN in water.

NaCN(aq)\rightarrow Na^+(aq.)+CN^-(aq)

e) The chemical equation for the reaction of NaCN and HCI

NaCN(aq)+HCl(aq)\rightarrow NaCl(aq.)+HCN(aq)

4 0
3 years ago
QUESTION 1
Svetlanka [38]

Answer:

C

Explanation:

Because nothing is left at the end of the reaction.

3 0
3 years ago
if a gas is moved from a large to a small container but its temperature in numbers of moles what would happen pressure of the ga
andreyandreev [35.5K]
The pressure will increase.

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