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2 years ago

Determine the osmotic pressure (atm) of an aqueous sugar solution with a

Chemistry

1 answer:

vaieri [72.5K]2 years ago

7 0

Answer:

61 atm

Explanation:

You can calculate osmotic pressure using the following equation:

π = i MRT

In this equation,

-----> π = osmotic pressure (atm)

-----> i = van't Hoff's factor

-----> M = molarity (M)

-----> R = Ideal Gas Constant (0.08206 atm*L/mol*K)

-----> T = temperature (K)

The van't Hoff's factor is the amount of ions the substance dissociates into. Since sugar does not dissociate, this value is 1. After converting Celsius to Kelvin, you can plug the values into the equation and simplify.

i = 1 R = 0.08206 atm*L/mol*K

M = 2.5 M T = 25 °C + 273.15 = 298.15 K

π = i MRT

π = (1)(2.5 M)(0.08206 atm*L/mol*K)(298.15 K)

π = 61 atm

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a) Por qué se humedeció la parte exterior del frasco?

b) Por qué el hielo disminuyó su volumen y ahora es agua?

c) Por qué puede haber agua en el exterior?

These are the three answers (in English).

First question:

a) Por qué se humedeció la parte exterior del frasco?

The question is Why did the outside of the bottle get wet?

Answer:

The outside of the bottle get wet because the ice cubes cooled the walls of the bottle, so the air surrounding the bottle also cooled.

The air contains humidity (water) in gas phase. The hotter the air the more the amount of humidity it can retain, the cooler the air the less the amount of humidity it can retain.

Then, when the air close to the walls of the bootle got cooler some of the water in the air became liquid and those are the drops of water that you see in the outside of the bottle.

Second question

b) Por qué el hielo disminuyó su volumen y ahora es agua?

The question is Why did the ice diminish its volume and now it is water?

Answer:

The ice diminished its volume and now it is water, becasue the ice, which is cooler than the surroundings, received heat energy (from the surroundings) and then its temperature increased. At some moment, this temperature reached the melting point of the ice (water) and it started to become liquid.

Third question

c) Por qué puede haber agua en el exterior?

The question is: Why can there be water outside?

Answer:

The water outside is outside since the beginning: it is in the air. You do not see it because it is gas state. When the air close to the walss of the bottle got cooler, part of the water in the air became liquid.

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3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

Electrolysis is used in the electroplating of metals. The same amount of current is passed through separate aqueous solutions of

Aleonysh [2.5K]

Answer:

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Explanation:

The valence of the elements in this case is as follows -

Cu - 2e-

Sn - 4e-

Cr - 3e-

CuSO4 cell will have the greatest amount of deposit among all three

The atoms of copper metal will deposit at the cathode. At the cathode, the least number of moles of electrons needed .

Hence, more amount of copper can be extracted out by the electrolyte

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3 years ago