Chemistry, 05.07.2020 14:01 queenjanet46

Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 2.74 oC compared to pure benzene. What is the molar mass of the unknown compound

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Answer from: NotYourStudent

The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.

With the addition of the solute to the solution, there has been a depression in the freezing point. The depression in the freezing point can be expressed as:

Depression in freezing point = Van't Hoff factor × Freezing point constant × molality

The molality can be defined as the moles of solute per kg solvent

Molaity = $\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}$

The depression in freezing point can be given as:

Depression in freezing point = Van't Hoff factor × Freezing point constant × $\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}$ ......(i)

Given, the depression in freezing point = 2.74 $\rm ^\circ C$

Van't Hoff factor = 1 (Molecular compound)

Freezing point constant (Kf) = 5.12 $\rm ^\circ C$ /m

Mass of solute = 26.7 g

Mass of solvent = 0.250 kg

Substituting the values in equation (i):

2.74 $\rm ^\circ C$ = 1 × 5.12

$\rm \dfrac{2.74}{5.12}$ = $\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{26.7}{0.250\;kg}$

0.535 = $\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;106.8$

Molecular mass of solute = $\rm \dfrac{106.8}{0.535}$ g/mol

Molecular mass of solute = 199.626 g/mol

The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.

For more information about the molality of the compound, refer to the link:

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Answer from: imsmart916

The molar mass of the unknown compound is 200 g/mol

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=2.74^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for molecular compound)

K_f = freezing point constant = 5.12^0C/m

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (benzene)= 0.250 kg

Molar mass of solute = M g/mol

Mass of solute = 26.7 g

$2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}$

M=200g/mol

Thus the molar mass of the unknown compound is 200 g/mol

Answer from: Quest

Cause its being chemically applied?

Answer from: Quest

Okay well gravitational energy is matter forced by gravity making energy though the kinetic energy then the electric energy produces electricity from the kinetic energy and the kinetic energy gets its energy from the gravitational energy. was that the answer you wanted or something else?

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Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f...

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