How many moles of gaseous boron trifluoride, BF3, are contained in a 4.3421 L bulb at 787.9 K if the pressure is 1.218 atm?

Answer:

15. Critical angle of glass and water combination, θ = 62.45°

16. Critical angle for the interface between Mystery A and glass, θ = 37.93°

Note; The question is incomplete. The complete question is as follows:

Medium Air Water Glass Mystery A Mystery B Table-2 Speed (m/s) 1.00 C 0.75 c 0.67 0.41 c 0.71 c n 1.00 1.33 1.50 Index of Refraction n of a given medium is defined as the ratio of speed of light in vacuum, c to the speed of light in a medium, v. n = c/v

Table-4: Incident Angle (degrees) Reflected Angle Refracted angle (degrees) (degrees) % Intensity of reflected ray 0 10 20 30 40 50 N/A N/A N/A 30 40 50 0 11.3 22.7 34.2 46.3 59.5 N/A N/A N/A 0.67 1.22 3.08 % Intensity of refracted ray 100 100 100 99.33 98.78 96.92

When rays travel from a denser medium to a less dense medium, we can define a critical angle of incidence θ such that refracted angle θ₂ = 90°. Applying Snell's law: Critical angle θ = sin-1(n₂/n₁).

When the angle of incidence is greater than the critical angle, 100% of the light intensity is reflected. This is called total internal reflection because all the light is reflected.

15. Calculate the critical angle of glass and water combination. Show your calculation.

16. What is the critical angle for the interface between Mystery A and glass?

Explanation:

15.  Applying Snell's law; Critical angle θ = sin-1(n₂/n₁).

where n₂,refractive index of water = 1.33, n₁, refractive index of glass = 1.50 since glass is denser than water

θ = sin-1(1.33/1.50)

θ = 62.45°

Critical angle of glass and water combination, θ = 62.45°

16.  Refractive index of mystery A , n = c/v

where v = 0.41 c

therefore, n = c / 0.41 c = 2.44

Critical angle for the interface between Mystery A and glass, θ = sin-1(n₂/n₁).

where n₂,refractive index of glass = 1.50, n₁, refractive index of mystery A = 2.44 since mystery A is denser than glass as seen from its refractive index

θ = sin-1(1.50/2.44)

θ = 37.93°

Critical angle for the interface between Mystery A and glass, θ = 37.93°

Answer:

The correct answer is: 0.064 V

Explanation:

In a oxidation-reduction reaction, the relation between Gibbs free energy (ΔGº) and cell potential (Eºcell) is given by:

[tex]\Delta G^{0} = -nFE^{0} _{cell}[/tex]

where n is the number of electrons that are transferred in the reaction and F is the Faraday constant (96,500 C/mol e-). Given: ∆G° = +18.55 kJ and n= 3 mol e-, we calculate Eºcell as follows:

+18.55 kJ = (-3 mol e-) x (96500 C/mol e-) x Eºcell

Eºcell= (+18.55 kJ)/(-3 mol e-) x (96500 C/mol e-)

Eºcell= (18550 J)/ (289500 C)

Eºcell= 0.064 J/C

Since 1 Volt= 1 Joule/1 Coulomb, thus:

Eºcell= 0.064 J/C = 0.064 V

Answer:

half life=0.693/rate constant =188.3

The half-life of this element is 188.32 years

The formula for calculating the half-life of an element is expressed according to the equation:

[tex]t_{1/2}=\frac{ln 2}{\lambda}[/tex]

Given the following parameters:

The rate constant for the decay = 3.68 × 10⁻³ day⁻¹.

Substitute into the formula for calculating the half-life as shown:

[tex]3.68\times 10^{-3}=\frac{0.693}{\lambda}\\ 0.00368=\frac{0.693}{\lambda}\\\lambda=\frac{0.693}{0.00368}\\\lambda = 188.32 years[/tex]

Hence the half-life of this element is 188.32 years

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Answer:

a. Methanol remains the same

b. Methanol decreases

c. Methanol increases

d. Methanol remains the same

e. Methanol increases

Explanation:

Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.

a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.

b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.

c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.

d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.

e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.

Answer:

The reaction is given as:

[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]

Explanation:

Bases are defined as those chemical substances which give hydroxide ions in their aqueous solutions.

[tex]BOH(s)\rightarrow B^+(aq)+OH^-(aq)[/tex]

When sodium hydroxide is added to water it gets dissociated into two ions that are sodium ions and hydroxide ions. Along with this heat energy also releases during this reaction.

The reaction is given as:

[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]

The equation to show how NaOH behaves in water is NaOH → Na⁺ + (OH)⁻

The compound that produce negative hydroxide (OH−) ions when dissolved

in water are called bases .

This compounds NaOH (sodium hydroxide) is an example of a base.

When it dissolves in water it dissociate to form  negative hydroxide (OH−)

ions  and positive sodium (Na+) ions.

It can be represented by the following equation:

NaOH → Na⁺ + (OH)⁻

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Answer: A

Explanation: Calculate the molar hydrogen ion concentration of each of the following biological solutions given the pH, Urine pH= 4.90

The given question is incomplete. The complete question is:

Compound X has a molar mass of 153.05 g/mol and the following composition:

element mass %

carbon 47.09%

hydrogen 6.59%

chlorine 46.33%

Write the molecular formula of X.

Answer: The molecular formula of X is [tex]C_6H_{10}Cl_2[/tex]

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 47.09 g

Mass of H = 6.59 g

Mass of Cl = 46.33 g

Step 1 : convert given masses into moles.

Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{47.09g}{12g/mole}=3.92moles[/tex]

Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.59g}{1g/mole}=6.59moles[/tex]

Moles of Cl =[tex]\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{46.33g}{35.5g/mole}=1.30moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{3.92}{1.30}=3[/tex]

For H = [tex]\frac{6.59}{1.30}=5[/tex]

For Cl =[tex]\frac{1.30}{1.30}=1[/tex]

The ratio of C : H: Cl= 3: 5 :1

Hence the empirical formula is [tex]C_3H_5Cl[/tex]

The empirical weight of [tex]C_3H_5Cl[/tex] = 3(12)+5(1)+1(35.5)= 76.5g.

The molecular weight = 153.05 g/mole

Now we have to calculate the molecular formula.

[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{153.05}{76.5}=2[/tex]

The molecular formula will be=[tex]2\times C_3H_5Cl=C_6H_{10}Cl_2[/tex]

Answer:

When balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.

Explanation:

Oxidation-reduction reactions or redox reactions are those in which an electron transfer occurs between the reagents. An electron transfer implies that there is a change in the number of oxidation between the reagents and the products.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The oxidation and reduction half-reactions, in a basic medium, adjust the oxygens and hydrogens as follows:

In the member of the half-reaction that presents excess oxygen, you add as many water molecules as there are too many oxygen. Then, in the opposite member, the necessary hydroxyl ions are added to fully adjust the half-reaction. Normally, twice as many hydroxyl ions, OH-, are required as water molecules have previously been added.

In short, you first adjust the oxygens with OH-, then you adjust the H with H₂O, and finally you adjust the charge with e-

So, when balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.

Answer:

c. balance the reaction as though under acidic conditions

Explanation:

When balancing redox reactions under basic conditions, a good technique is to first balance the reaction as though under acidic conditions. We then adjust the result to reflect the basic conditions.

Answer:

Heat of vaporization refers to condensation and heat of fusion refers to melting.

Explanation:

Heat of vaporization or heat of evaporation, is defined as the energy required to convert liquid substance into a gas which creates condensation. As a gas condenses to a liquid, heat is released.

Heat of fusion refers to melting because heat of fusion is defined as the energy required to change any amount of substance when it melts.

Hence, the correct answer is "Heat of vaporization refers to condensation and heat of fusion refers to melting.".

Answer:

Volume of the calcium hydroxide solution used is 0.0235 mL.

Explanation:

Moles of KHP =

According to reaction, 2 moles of KHP  with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;

of calcium hydroxide

Molarity of the calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

Volume of the calcium hydroxide solution used is 0.0235 mL.

Answer:

[tex]Kp=5.14[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]N_2+3H_2\rightarrow 2NH_3[/tex]

Thus, the equilibrium expression is written as:

[tex]Kp=\frac{p_{NH_3}^2}{p_{N_2}p_{H_2}^3}[/tex]

And in terms of the reaction extent:

[tex]Kp=\frac{(2x)^2}{(10-x)(10-3x)^3}[/tex]

Thus, from the equilibrium pressure of ammonia we can compute the reaction extent:

[tex]p_{NH_3}=2x=6.0 atm\\\\x=3.0atm[/tex]

Therefore, the equilibrium constant turns out:

[tex]Kp=\frac{(2*3.0)^2}{(10.0-3.0)(10.0-3*3.0)^3}\\\\Kp=5.14[/tex]

Regards.

Answer:

Bakelite is a polymer made up of the monomers phenol and formaldehyde. This phenol-formaldehyde resin is a thermosetting polymer.

Answer: The monomers of bakelite are formaldehyde and phenol

Explanation:

Answer:

OPTION C is correct

C) 1.07 g

Explanation:

CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)3(aq)

But we know molarity molarity= number of moles of solute/ volume of the solution

M= n/V

From the equation above

number of moles of Cacl2 = (37.5 ×0 .100 × 10^-3) = 0.00375 moles.

Then

1 mole of Cacl2 yields 2 moles of Agcl2

0.00375 moles of Cacl2 will produce let say Y.

Y= (0.00375 ×2)/1

= 0.0075 moles.

Number of moles of Agcl2 = mass /molar mass of Agcl

Molar mass of Agcl = 178

Then mass = 178 ×0.0075 = 1.047

Therefore, the mass of agcl precipitate is

1.07 g

Answer:

The best reducing agent is Zn(s)

Explanation:

A reducing agent must to be able to reduce another compound, by oxidizing itself. Consequently, the oxidation potential must be high. The oxidation potential of a compound is the reduction potential of the same compound  with the opposite charge. Given the reduction potentials, the best reducing agent will be the compound with the most negative reduction potential. Among the following reduction potentials:

Cd₂⁺(aq)|Cd(s) ⇒ -0.40 V

Zn²⁺(aq)|Zn(s) ⇒ -0.76 V

Ni²⁺(aq)|Ni(s) ⇒‑0.25 V

Cu²⁺(aq)|Cu(s) ⇒ +0.34 V

Ag⁺(aq)|Ag(s) ⇒ +0.80 V

The most negative is Zn²⁺(aq)|Zn(s) ⇒ -0.76 V

From this, the most reducing agent is Zn. Zn(s) is oxidized to Zn²⁺ ions with the highest oxidation potential (0.76 V).

Answer:

atoms contain tiny negatively charged subatomic particles or electrons

Answer:

Option D. A lipid with three unsaturated fatty acid.

Explanation:

The molecule in the diagram above contains three fatty acid.

A careful observation of the molecule reveals that each of the three fatty acids contains a double bond.

The presence of a double bond in a compound shows that the compound is unsaturated.

Thus, we can say that the molecule is a lipid with three unsaturated fatty acid.

Answer:

CH2Cl2 is polar

Explanation:



Answer:

Underneath the frozen upper layer, the water remains in its liquid form and does not freeze. Also, oxygen is trapped beneath the layer of ice. As a result, fish and other aquatic animals find it possible to live comfortably in the frozen lakes and ponds. ... This irregular expansion of water is called anomalous expansion.

Explanation:

so pretty much its because there is water under the top frozen layer and air is trapped underneath the ice

Answer:

1.6x10¹¹ = Kc

Explanation:

For the reaction:

AgCl(s) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) + Cl⁻(aq)

Kc is defined as:

Kc = [Ag(CN)₂⁻] [Cl⁻] / [CN⁻]²

Ksp of AgCl is:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

Where Ksp is:

Ksp = [Ag⁺] [Cl⁻] = 1.6x10⁻¹⁰

In the same way, Kf of Ag(CN)₂⁻ is:

Ag⁺(aq) + 2CN⁻ ⇄ Ag(CN)₂⁻

Kf = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = 1.0x10²¹

The multiplication of Kf with Ksp gives:

[Ag⁺] [Cl⁻] *  [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = Ksp*Kf

[Ag(CN)₂⁻] [Cl⁻] / [CN⁻]² = Ksp*Kf

Obtaining the same expression of the first reaction

That means Ksp*Kf = Kc

1.6x10⁻¹⁰*1.0x10²¹ = Kc

Answer:

[tex]pH=10.45[/tex]

Explanation:

Hello,

In this case, for the dissociation of the given base, we have:

[tex]base\rightleftharpoons OH^-+CA[/tex]

Whereas CA accounts for conjugated acid and OH⁻ for the conjugated base. In such a way, equilibrium expression is:

[tex]Kb=\frac{[OH^-][CA^+]}{[base]}[/tex]

And in terms of the reaction extent [tex]x[/tex] we can write:

[tex]1.6x10^{-6}=\frac{x*x}{0.05M-x}[/tex]

For which the roots are:

[tex]x_1=-0.000284M\\x_2=0.000282M[/tex]

For which clearly the result is the positive root which also equals the concentration of hydroxyl ions and we can compute the pOH:

[tex]pOH=-log([OH^-])=-log(0.000282)\\\\pOH=3.55[/tex]

And the pH:

[tex]pH=14-pOH=14-3.55\\\\pH=10.45[/tex]

Regards.

The pH of the solution is 10.45.

Let us represent codeine with the generic formula BH. We can set up the ICE table as follows;

              :B(aq) + H2O(l) ⇄ BH(aq)  + OH^-(aq)

I            0.05                        0                0

C           -x                            +x                +x

E        0.05 - x                      x                  x

We know that the Kb of codeine is 1.6 x 10^-6, Hence;

1.6 x 10^-6 = x^2/0.05 - x

1.6 x 10^-6 (0.05 - x ) =  x^2

8 x 10^-8 - 1.6 x 10^-6x =  x^2

x^2 +  1.6 x 10^-6x - 8 x 10^-8 = 0

x = 0.00028 M

The concentration of hydroxide ions = 0.00028 M

Given that pOH = - log[0.00028 M]

pOH = 3.55

pH + pOH = 14

pH = 14 - 3.55

pH = 10.45

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Answer:

13.11 g.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below :

N2 + O2 —> 2NO

Next, we shall determine the mass of O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 1 x 32 = 32 g.

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO from the balanced equation = 2 x 30 = 60 g.

From the balanced equation above,

32 g of O2 reacted to produce 60 g of NO.

Finally, we shall determine the theoretical yield of NO as follow:

From the balanced equation above,

32 g of O2 reacted to produce 60 g of NO.

Therefore, 6.99 g of O2 will react to produce = (6.99 x 60)/32 = 13.11 g of NO.

Therefore, the theoretical yield of nitrogen monoxide, NO is 13.11 g.

Answer:

D. Rare

Explanation:

Qualitative research has to do with non-numerical data and is used to understand the beliefs of a group of people which can be gotten from surveys, questionnaires, interviews, etc.

A qualitative researcher may select a RARE case that is unusually rich in information pertaining to the research question.

This is because he wants to get an insight into the why, how, where and when of that particular ase as it's not a usual occurrence.

answer is oxygen .

oxygen is an exception in octet rule

Among the following given elements,oxygen is an element which cannot have an expanded octet.

Expanded octet is a condition where an octet has more than 8 electrons and which is called as hyper-valency state. This concept is related to hybrid orbital theory and Lewis theory. Hyper-valent compounds  are not less common  and are of equal stability as the compounds which obey octet rule.

Expansion of octet is possible for elements from third period on wards only as they have low-lying empty d - orbitals which can accommodate more than eight electrons.

Expanded octet is not applicable to oxygen as it is second period of periodic table and has less than ten electrons and even does  not have the 2d -orbitals   due to which it  does not fulfill the criteria of an element to have an expanded octet.

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Answer:

2Cl——>Cl2+2e-

Explanation:

It shows an electron loss or gain

Answer:

[tex]6.4knips[/tex]

Explanation:

Hello,

In this case, given the stated equivalences, we can use the following proportional factor in order to compute the required knips:

[tex]knips=1.00knop*\frac{6bippy}{1knop} *\frac{8pringle}{10bippy}* \frac{3blip}{18pringle} *\frac{4clips}{5blips} *\frac{10knip}{1clip} \\\\=6.4knips[/tex]

Regards.

1 knop=6.4 knips

First convert knop to bippy:-

[tex]1\ knop\times\frac{6\ bippy}{1\ knop} =6\ bippy[/tex]

Now, Convert 6 bippy to pringle:-

[tex]6\ bippy\times\frac{8\ pringle}{10\ bippy} =4.8\ pringle[/tex]

Now, convert 4.8 pringle to blip:-

[tex]4.8\ pringle\times\frac{3\ blip}{18\ priangle} =0.8\ blip[/tex]

Now, convert 0.8 blip to clips as follows:-

[tex]0.8\ blip\times\frac{4\ clips}{5\ blip} =0.64\ clip[/tex]

Now, convert 0.64 clip to knips:-

[tex]0.64\ clip\times\frac{10\ knip}{1\ clip} =6.4\ knip[/tex]

Hence, the following conversion is as follows:-

1.00 knop=6.4 knips

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Answer:

[H₃O⁺] = 2.5 × 10⁻¹³ M

pH = 12.6

Explanation:

Step 1: Given data

Concentration of OH⁻: 0.04 M

Step 2: Calculate the concentration of H₃O⁺

Let's consider the self-ionization of water reaction.

2 H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)

The ionic product of water is:

Kw = [OH⁻] × [H₃O⁺] = 10⁻¹⁴

[H₃O⁺] = 10⁻¹⁴ / [OH⁻]

[H₃O⁺] = 10⁻¹⁴ / 0.04

[H₃O⁺] = 2.5 × 10⁻¹³ M

Step 3: Calculate the pH

The pH is:

pH = -log [H₃O⁺] = -log 2.5 × 10⁻¹³ = 12.6

Explanation:

According to your question,

no. a. ans would be like; chemical formula is defined as the an expression which determines no. and type of molecule of a compound.

b. no. ans; it tells that what type of compound is formed with the type and no. of atoms present in the atom.

c. no ans; the formulation of h2so4 states that it is acid named as hydrochloric acid which is formed by reacting of hydrogen (2 atoms ) ,sulpher (*1atom) and oxygen(4atoms).

Hope it helps...

Answer:

315 g

Explanation:

Step 1: Write the thermochemical equation

2 H₂O(l) → 2 H₂(g) + O₂(g)      ΔH = +572 kJ

Step 2: Calculate the molar of water decomposed by 5.00 × 10³ kJ of energy

According to the thermochemical equation, 572 kJ are required to decompose 2 moles of water.

5.00 × 10³ kJ × (2 mol/572 kJ) = 17.5 mol

Step 3: Calculate the mass corresponding to 17.5 moles of water

The molar mass of water is 18.02 g/mol.

17.5 mol × 18.02 g/mol = 315 g

Answer:

ΔG°  = 1022. 8 kJ

Explanation:

ΔH° = –199 kJ/mol

ΔS° = –4.1 J/K·mol

T = 25°C = 25 + 273 = 298K (Converting to kelvin temperature)

ΔG° = ?

The relationship between these varriables are;

ΔG° = ΔH°  - TΔS°

ΔG° = –199 - 298 (–4.1)

ΔG° = -199 + 1221.8

ΔG°  = 1022. 8 kJ

To calculate ΔG, the following equation should be used:

ΔG° = ΔH°  - TΔS°

Given:

ΔH° = –199 kJ/mol,  

ΔS° = –4.1 J/K·mol

T=25+273K=298K  

Substitute the respective values:

ΔG° = ΔH°  - TΔS°

       =–199 kJ/mol-(298K×(-4.1 J/K·mol*(1KJ/1000J)))

       =-197.78kJ/mol

Thus, we can conclude the value of ΔG°=-197.78kJ/mol  

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Answer:

494.49 g of NaCl.

Explanation:

Data obtained from the question include the following:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mass of NaCl =.?

Next, we shall determine the number of mole of NaCl in the solution.

Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as

Molality = mole of solute /Kg of solvent

With the above formula, we can obtain the number of mole NaCl in the solution as follow:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mole of NaCl =..?

Molality = mole of solute /Kg of solvent

3.140 = mole of NaCl /2.692

Cross multiply

Mole of NaCl = 3.140 x 2.692

Mole of NaCl = 8.45288 moles

Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:

Mole of NaCl = 8.45288 moles

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =.?

Mole = mass /Molar mass

8.45288 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 8.45288 × 58.5

Mass of NaCl = 494.49 g.

Therefore, 494.49 g of NaCl are present in the solution.

The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.

Molality can be defined as the mass of solute present in 1000 grams of solvent.

Molality = [tex]\rm \dfrac{moles\;of\;solute}{mass\;of\;solvent\;(kg)}[/tex]

The moles of NaCl has been calculated as;

3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{mass\;of\;water\;(kg)}[/tex]

3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{2.692\;kg}[/tex]

Moles of NaCl = 3.140 [tex]\times[/tex] 2.692

Moles of NaCl = 8.45288 mol

The moles can be expressed as;

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Molecular weight of NaCl = 58.5 g/mol

The mass of NaCl can be calculated as:

8.45288 mol = [tex]\rm \dfrac{mass\;of\;NaCl}{58.5\;g/mol}[/tex]

Mass of NaCl = 58.5 g/mol [tex]\times[/tex] 8.45288 mol

Mass of NaCl = 494.493 grams.

The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.

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