A block solution of acid 70.5% by mass HCIO_4 and 29.576 by water Then density o

Question

A block solution of acid 70.5% by mass HCIO_4 and 29.576 by water Then density of the solution is 1.57 at 20 degree C. What is the initially of the solution?_________12.5 M 11.7 M 10.5M 5.5M The atmospheric pressure on Pluto is approximately 0.3pa. Convert this pressure to atmospheres [1 atm = 101325Pa]______3 times 10^4 atm 0.08 atm 3 times 10^4 atm 0.3 atm A sample of oxygen occupies44.40 mL at 1.20 atm what will the volume of the oxygen if the pressure to 0.80 atm at a constant temperature ?___________0.022 ml. 67 mL 43 mL 29 mL A sample of gas has a volume of 3.80 L at an unknown temperature. When the sample is placed into a cold bath at -5.00dgree C, its volume decreases to 1.12 L. What was the initial temperature of the sample,in degree C?_______-194 degree C 5.00 degree C 636 degree C 909 degree C A cylinder fitted with a piston contains 0 333 moles of gas with a volume of 8271L Determine the total volume of gas In the cylinder after 0.187 moles of gas are added . (Assume that the temperature and pressure are held constant.)_________16 2 L 4.64 L 12 9L 3 82 L Calculate the volume of 0,222 motes of gas at a pressure of 1.30 atm and a temperature of 298 K________0.0509 L 4.18 L 0.239 L 88.0 L A sample of NO_2gas has a density of 2 25 g L^-1at 277 K What is the pressure of the gas. in bar? [R = 0.0831447 L bar mol^-1 K^-2]________2 22 bar 1.11 bar 1.13 bar 0.901 bar A gas mixture has a total pressure of 774 Torr The mixture contains 56 0 Tort O_2 145 Torr N_2, 322 Torr He. and some fluorine gas. What is the mote fraction o1 the oxygen gas in this mixture?__________0.167 0.416 0.324 0.0724

Explanation / Answer

HClO4 = 70.5% m/m

Water= 29.5 % m/m

Density= 1.67 g/mL

Molarity of HClO4 = ?

Since we know density and knowing that 1.67g are present in 1 mL of solution, we calculate grams present in 1L (1000 mL) of solution. We do this since we need to calculate Molarity, and Molarity is equal to mols/L, in this case will be mols per liter of solution.

1000mL (1.67 g / 1 mL) = 1670 g

Which means that 1L of solution will have a mass of 1670 g.

Now we calculate mass of HClO4:

We know that the percent of HClO4 in the sample is 70.5%. It means that we have 70.5g of HClO4 per 100 g of solution.

70.5% (1670g / 100%) = 1177.35 g of HClO4

Now we calculate mols of HClO4, we do this with the molar mass of HClO4 (100.46g/mol). Since 1 mol of HClO4 has 100.46 g of HClO4, how many mols will represent 1177.35g of HClO4?

Molar mass HClO4 = 1.008g/mol + 35.453g/mol + (15.9994g/mol x 4) = 100.46 g/mol

1177.35g HClO4 (1 mol / 100.46 g HClO4) = 11.72 mols of HClO4

Since we based our calculation in 1L of solution. And Molarity represent mols/L

Molarity of HClO4 = 11.72 mols / 1 L

Molarity of HClO4 = 11.72 M

We need to convert the atmospheric pressure in pascals into atmospheres.

Since 1 atm = 101325 Pa

0.3 Pa (1 atm / 101325 Pa) = 3 x 10-6 atm

Here we need to use the formula P1V1 = P2V2. This formula is from Boyle’s Law, which states that: (initial pressure) x (initial volume) = (final pressure) x (final volume)

We have that:

P1 = 1.20 atm

V1= 44.40 mL

P2 = 0.80 atm

V2 = ?

So, we rearrange equation:

V2 = P1V1 / P2

V2 = 44.40mL x 1.20 atm / 0.80 atm

V2 = 66.6 mL = 67.0 Ml

Since in this problem we do not have any mention of pressure, we assume ideal gas conditions. And we have to use Charles’s Gas Law:

We have: V1 / T1 = V2 / T2

Where, (V1= initial volume) / (T1= initial temperature) =    (V2= final volume) / (T2= final temperature)

Converting Celsius into Kelvins: 273.15 + (-5.00 °C) = 268.15 K

Substituting for values:

3.80 L / T1 = 1.12 L / 268.15 K

3.80 L / T1 = 0.0042

T1 = 3.80 L /0.0042

T1 = 909.79 K

°C = 909.79 K – 273.15 K = 636.64 °C

Here we need to use Avogadro’s Law:

V1/n1 = V2/n2

Where,

V1= initial volume

n1= initial number of mols

V2= final volume

n2= final number of mols

We have that:

V1= 8.27 L

n1= 0.333 mols

n2= 0.187 mols

V2= ?

Substituting in equation:

8.27 L / 0.333 mols = V2/0.187 mols

V2= 24.83 x 0.187

V2= 4.64 L

Here we need Ideal Gas Law equation:

PV = nRT

Where:

P= pressure ------> (1.30 atm)

V= volume -----> (?)

n= number of mols of substance ------> (0.222 mols)

R= gas constant (in this case we use it in atm: 0.0821 L . atm / K . mol)

T= temperature (in Kelvins) -----> (298K)

We rearrange equation to solve for Volume, and put values on it:

V = n R T / P

V= (0.222 mols) x (0.0821 L . atm / K . mol) x (298 K) / 1.30 atm

V = 4.18 L

Here we need Ideal Gas Law equation, similar to previous question:

PV = nRT

Where:

P= pressure ------> (?)

V= volume

n= number of mols of substance

R= gas constant (in this case we use it in bars: 0.0831447 L bar / mol K)

T= temperature (in Kelvins) -----> (277K)

With density and molar mass we can calculate mols of NO2.

With density we know that 2.25 g are present in 1 L of solution.

Molar mass of NO2 = 46.0055 g/mol

Mols of NO2 = 2.25 g (1 mol / 46.0055 g) = 0.0489 mols of NO2

We know that we have grams per liter, so, volume will be 1 L.

We rearrange equation to solve for Pressure, and put values on it:

P = n RT / V

P= (0.0489 mols) x (0.0831447 L bar / mol K) x (277 K) / 1 L

P = 1.1262 bars = 1.13 bars

The number of mols of each gas present in the mixture is proportional to its partial pressure. So, with this we can calculate mole fraction of oxygen gas.

Mol Fraction of O2 = Pressure of O2 in the mixture / Total Pressure

Mol Fraction of O2 = 56.0 Torr / 774 Torr

Mol Fraction of O2 = 0.0724 Torr

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