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3 years ago

8

How could you show experimentally that the molecular formula of propene is c3h6, not ch2?

2 answers:

Valentin [98]3 years ago

6 0

Answer : By performing the experiment for calculating the elevation in boiling point and finding out the <span>ebuilloscopic</span> constant for the solvent we can find the molecular formula of the unknown solute.

Also by performing the experiment for depression in freezing point one can find the molecular formula for any unknown solute. We can find the cryoscopic constant for the solute and it will help in determining the molecular formula.

katen-ka-za [31]3 years ago

5 0

Answer:

By performing experiments related with the elevation in boiling point and depression in freezing point.

Explanation:

Hello,

In this case, we use the concept of colligative properties of solutions because they depend on the concentration of solute molecules or ions rather than its identity. They are classified as vapor pressure lowering, boiling point elevation, freezing point depression and osmotic pressure.

Elevation in boiling point and depression in freezing point are suitable for us to determine the molecular formula of propane as shown below:

- Elevation in Boiling Point: by performing an elevation in boiling point experiment, it is possible to quantify ebuilloscopic constant for the solvent, in order words, the molal (referred to molality) boiling point elevation constant of propane, using it as a solute. Then, one applies the formula:

$\Delta T_b=i*Kb*m_{solute}$

To compute the molality and subsequently propane's molecular mass.

- Depression in freezing point: by performing this one, it is possible to determine the molecular mass of the propane. Thus, we find the molal freezing point depression constant of the solvent to subsequently apply the shown below formula:

$\Delta T_f=i*Kf*m_{solute}$

Which is suitable to compute the molality as well as in the previous case.

Best regards.

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Two cylinders are made of the same material. Cylinder A is one-fourth (1/4) the length of cylinder B and it has a radius that is

tankabanditka [31]

Answer:

$M_{A}$ : $M_{B}$ = 4 : 1

Explanation:

Given that:

Volume of a cylinder = $\pi$ $r^{2}$ h

For cylinder A, $l_{A}$ = $h_{A}$ = $\frac{1}{4} l_{B}$ and $r_{A}$ = 4 $r_{B}$ .

Volume of cylinder A = $\pi$ $(4r_{B}) ^{2}$ × $\frac{1}{4} l_{B}$

= 4 $\pi$ $r_{B} ^{2}$ $l_{B}$

Volume of cylinder B = $\pi$ $(r_{B}) ^{2}$ $l_{B}$

= $\pi$ $r_{B} ^{2}$ $l_{B}$

To determine the ratio of their masses, density (ρ) is defined as the ratio of the mass (M) of a substance to its volume (V).

i.e ρ = $\frac{M}{V}$

Thus, since the cylinders are made from the same material, they have the same density (ρ). So that;

density of A = density of B

density of A = $\frac{M_{A} }{4\pi r_{B} ^{2}l_{B} }$

density of B = $\frac{M_{B} }{\pi r_{B} ^{2}l_{B} }$

⇒ $\frac{M_{A} }{4\pi r_{B} ^{2}l_{B} }$ = $\frac{M_{B} }{\pi r_{B} ^{2}l_{B} }$

The ratio of mass of cylinder A to that of B is given as;

$\frac{M_{A} }{M_{B} }$ = $\frac{4\pi r_{B} ^{2} l_{B} }{\pi r_{B} ^{2} l_{B} }$

⇒ $\frac{M_{A} }{M_{B} }$ = $\frac{4}{1}$

Therefore, $M_{A}$ : $M_{B}$ = 4 : 1

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