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Ugo [173]
3 years ago
15

The osmotic pressure exerted by a solution is equal to the molarity multiplied by the absolute temperature and the gas constant

R. Suppose the osmotic pressure of a certain solution is measured to be 5.1 atm at an absolute temperature of 312 K. Write an equation that will let you calculate the molarity of this solution. Your equation should contain only symbols. Be sure you define each symbol other than R .
Chemistry
1 answer:
enot [183]3 years ago
4 0

Answer: C=\frac{\pi}{R\times T}

Explanation:-

Osmotic pressure is a colligative property which depends on the amount of solute added.

\pi=CRT

\pi = osmotic pressure = 5.1 atm

C= concentration in Molarity

R= solution constant = 0.0821 Latm/Kmol

T= temperature = 312 K

C=\frac{\pi}{R\times T}

C=\frac{5.1atm}{0.0821Latm/Kmol\times 312K}=0.2mol/L

Thus the equation for calculating the molarity of this solution is C=\frac{\pi}{R\times T}

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Write a sentence that descrices how to determine the number of moles of a compound in known mass of a
sattari [20]

Explanation:

<u>Moles is denoted by given mass divided by the molecular mass ,  </u>

Hence ,  

n = w / m

n = moles ,  

w = given mass ,  

m = molecular mass .

For example ,

For  a compound X ,

The given mass i.e. w = 20 g

and the molecular mass ,i.e. , m = 10 g / mol

Then the moles can easily be calculated by using the above formula ,

n = w / m

n = 20 g / 10 g/mol = 2 mol

Hence , answer = 2 mol.

6 0
3 years ago
The number of energy levels to which an electron can jump depends on the
Kitty [74]
The number of energy levels to which an electron can jump depends on the amount of energy the electron possesses. Each energy level has a specific amount of energy an electron needs to have before it can be in there. So, if an electron doesn't have enough energy to be in that energy level then it won't jump to that higher level.
6 0
3 years ago
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Picture of gas laws question below:
Vera_Pavlovna [14]

Volume of the tank is 5.5 litres.

Explanation:

mass of the CO2 is given 8.6 grams

Pressure of the gas is 89 Kilopascal which is 0.8762 atm

Temperature of the gas is 29 degrees ( 0 degrees +273.5= K) so (29+273)

R = gas constant 0.0821 liter atmosphere per kelvin)

FROM THE IDEAL GAS LAW

PV=nRT ( P Pressure, V Volume, n is number of moles of gas, R gas constant, Temperature in Kelvin)

no of moles = mass/atomic mass

                    =  8.6/44

                    = 0.195 moles

now putting the values in equation

V=nRT/P

  = 0.195*0.0821*302/ 0.8762

  = 5.5 litres.

As the carbon dioxide gas occupies the volume os the tank hence volume of tank is 5.5 litres.

4 0
3 years ago
Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a
nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

               Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

  • n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

                             =  (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

                                                                                           = 75.3%

4 0
3 years ago
A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

3. Molal concentration of EG

b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

7 0
3 years ago
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