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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answered: 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:

link

answered: 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

answered: Guest

a rain shadow is a dry region of land on the side that is protected against the prevailing winds of a mountain range. the prevailing winds are the winds that occur most of the time in a particular place of the earth

prevailing winds carry air toward the mountain range. as the air rises up over a mountain range, the air cools, water vapor condenses, and clouds form. on this side of the mountains, called the windward side, precipitation falls in the form of rain or snow. this dry air produces a rain shadow.

answered: Guest

the correct answer is:

a. formation of precipitation