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aleksandr82 [10.1K]
3 years ago
14

A 0.050 MM solution of AlCl3 had an observed osmotic pressure of 3.85 atmatm at 20 ∘C . Calculate the van't Hoff factor iii for

AlCl3.
Chemistry
1 answer:
alekssr [168]3 years ago
8 0

Answer:

The van't Hoff factor = 3.20

Explanation:

Step 1: data given

Osmotic pressure of a 0.050 M Solution is 3.85 atm

Temperature = 20.0 °C

Step 2:

Osmotic pressure depends on the molar concentration of the solute but not on its identity.

We can calculate the osmotic pressure by:

π = i.M.R.T

 ⇒ with π = osmotic pressure = 3.85 atm

 ⇒ with i = van 't Hoff factor  = TO BE DETERMINED

 ⇒ with M = molar concentration of the solution =0.050 M

 ⇒ with R = gas constant =0.08206 L * atm / mol* K)

  ⇒ with T = Temperature of the solution =20°C = 293 K

i = π / M.R.T

i = 3.85 / 0.050*0.08206*293

i = 3.20

The theoretical Van't Hoff factor is 4:

AlCl3(aq) → Al^3+(aq) + 3Cl^-(aq)

AlCl3 dissociates in 1 mol Al^3+ + 3 moles Cl-

Due to the interionic atractions the Van't hoff factor is less than the theoretical value of 4

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3 years ago
Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How ma
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Answer: 12.0 milliliters of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn.

Explanation:

moles =\frac{\text {given mass}}{\text {Molar mass}}

moles of zinc =\frac{2.55g}{65.38g/mol}=0.0390moles

The balanced chemical equation is :

Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)

According to stoichiometry:

1 mole of zinc reacts with = 2 moles of HCl

Thus 0.0390 moles of zinc reacts with = \frac{2}{1}\times 0.0390=0.0780 moles of HCl

To calculate the volume for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution in ml}}     .....(1)

Molarity of HCl solution = 6.50 M

Volume of solution = ?

Putting values in equation 1, we get:

6.50M=\frac{0.0780\times 1000}{\text{Volume of solution in ml}}

{\text{Volume of solution in ml}}=12.0ml

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Number of grams of hcl that can react with 0.500 grams of Al(OH)3
vlabodo [156]

Answer:

0.7g of HCl

Explanation:

First, let us write a balanced equation for the reaction between HCl and Al(OH)3.

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Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:

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Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g

Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:

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From the equation above,

78g of Al(OH)3 reacted with 109.5g of HCl.

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