Chemistry, 28.07.2021 04:40, amcda213040

Suppose that you add 24.3 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 3.14 oC compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound

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answered: jtorres0520

Solution :

We know that :

$\Delta T_f = k_f.m$ and $m=\frac{w_2}{m_2 \times w_1}$

Then, $\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$ ..................(1)

Where,

w_1 = amount of solvent (in kg)

w_2 = amount of solute (in kg)

m_2 = molar mass of solute (g/mole)

= molality of solution (mole/kg)

Given :

$\Delta T_f$ = $3.14\ ^\circ C$ , $k_f= 5.12\ ^\circ C/m$

$=5.12 \ ^\circ C/mole/kg$

$=5.12 \ ^\circ C \ kg/mole$

w_1 = 0.250 kg, w_2 = 24.3 g

Then putting this values in the equation is (1),

$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$

$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$

m_2= 158.49 g/mole

So, the molar mass of the unknown compound is 158.49 g/mole.

answered: Guest

that's just how they are

explanation:

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answered: Guest

i do not get it at all

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Suppose that you add 24.3 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f...