The osmotic pressure, π, of a solution of glucose is 132 atm . find the molarity of the solution at 298 k.


Answer 1

The formula for osmotic pressure is:

\Pi = iMRT

where \Pi is osmotic pressure, i is van't Hoff's factor, M molarity, R is Ideal gas constant, and T is Temperature.

\Pi = 132 atm

The van't Hoff's factor for glucose, i = 1

R = 0.08206 Latmmol^(-1)K^(-1)

T = 298 K

Substituting the values in the above equation we get,

132 atm = 1* M* 0.08206 Latmmol^(-1)K^(-1)* 298

M = (132 atm)/(1* 0.08206 Latmmol^(-1)K^(-1)* 298) = 5.4797 molL^(-1) \simeq 5.48 molL^(-1)

So, the molarity of the solution is 5.48 molL^(-1).

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Calculate the molality of a 35.4 % (by mass) aqueous solution of phosphoric acid (H3PO4) (35.4 % means 35.4 g of H3PO4in 100 g of solution)



3.6124 m/kg


Molality is calculated as moles of solute (mol) divided by kilogram of solvent (kg). Here, we can find these numbers by using the 35.4%, which gives us  35.4 g of H3PO4 and 100 g of solution to work with.

To go from grams to moles for the phosphoric acid, you need to find the molar mass of the compound or element and divide the grams of the compound or element by that molar mass.

Here, the molar mass for phosphoric acid is 97.9952 g/mol. The equation would look like this:

35.4 g x 1 mol / 97.9952 g = 0.3612422 mol

Next, the 100 g of solvent can easily be converted to 0.1 kg of solvent.

To find the molality, divide the moles of solute and kilograms of solution.

0.3612422 mol / 0.1 kg = 3.6124 m/kg

At 25.0 ⁰C the henry's law constant for hydrogen sulfide(H2S) gas in water is 0.087 M/atm. Caculate the mass in grams of H2S gas that can be dissolved in 400.0 ml of water at 25.00 C and a H2S partial pressure of 2.42atm.


Answer: The mass of hydrogen sulfide that can be dissolved is 2.86 grams.


Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the gas.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_(H_2S)=K_H* p_(liquid)


K_H = Henry's constant = 0.087M/atm

p_(H_2S) = partial pressure of hydrogen sulfide gas = 2.42 atm

Putting values in above equation, we get:

C_(H_2S)=0.087M/atm* 2.42atm\n\nC_(H_2S)=0.2105M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = 0.2105 M

Molar mass of hydrogen sulfide = 34 g/mol

Volume of solution = 400.0 mL

Putting values in above equation, we get:

0.2105M=\frac{\text{Mass of hydrogen sulfide}* 1000}{34g/mol* 400.0mL}\n\n\text{Mass of }H_2S=(0.2105* 34* 400)/(1000)=2.86g

Hence, the mass of hydrogen sulfide that can be dissolved is 2.86 grams.

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A 0.7 ft diameter hole forms in a tank containing butane at 19 atmg and 76 degrees Fahrenheit. Determine the maximum possible mass flow rate through this leak in lb m / min, if the external pressure is 1 atm.



Q = 3,534.4 lbm/s = 212,062 lbm/min


Mass flowrate of discharge or leakage mass flowrate (Q) is given as

Q = AC₀√(2ρgP)

A = Cross sectional Area of leakage = (πD²/4) = (π×0.7²)/4

A = 0.385 ft²

C₀ = discharge coefficient = 0.98 (For maximum discharge flow rate, the flow is turbulent with discharge coefficient within 1% of 0.98)

ρ = density of butane at 76°F = 35.771 lbm/ft³

g = acceleration due to gravity = 32.2 lbm.ft/lbf.s²

P = Gauge Pressure in the tank = (absolute pressure) - (external pressure) = 19 - 1 = 18 atm = 38091.9 lbf/ft²

Q = AC₀√(2ρgP)

Q = (0.385)(0.98)√(2×35.771×32.2×38091.9)

Q = 3,534.4 lbm/s = 212,062 lbm/min

Hope this Helps!!!

Write the balanced reaction and solubility product expression (KSP) for dissolving silver chromate: Ag2CrO4(s). Include all charges, stoichiometric coefficients, and phase subscripts.



2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓

Ksp = [2s]²  . [s] → 4s³


Ag₂CrO₄ → 2Ag⁺  + CrO₄⁻²

Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓ Ksp

That's the expression for the precipitation equilibrium.

To determine the solubility product expression, we work with the Ksp

Ag₂CrO₄ (s)  ⇄ 2Ag⁺ (aq)  + CrO₄⁻² (aq)   Ksp

                          2 s                 s

Look the stoichiometry is 1:2, between the salt and the silver.

Ksp = [2s]²  . [s] → 4s³