liluv6650

liluv6650

10.09.2021 •

Chemistry

What quantity in moles of potassium atoms do you have if you have 3.50 x 10^22 atoms of potassium.

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Ответ:

birdman37361

07.10.2021

Answer : The volume of ethylene glycol is, 3.92 liters.

The boiling point of a solution is, 105.6^oC

Explanation :

First we have to calculate the molality of ethylene glycol.

Formula used :

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

where,

$\Delta T_f$ = change in freezing point = 20^oC

i = Van't Hoff factor = 1 (for non-electrolyte)

K_f = freezing point constant = 1.86^oC/m

m = molality of ethylene glycol = ?

Now put all the given values in the above formula, we get:

$20^oC=1\times 1.86^oC/m\times m$

m=10.8mol/kg

Now we have to calculate the mass of water.

$\text{Mass of water}=\text{Volume of water}\times \text{Density of water}$

$\text{Mass of water}=6.50L\times 1kg/L=6.50kg$

Now we have to calculate the moles of ethylene glycol.

Moles of ethylene glycol = Molality × Mass of water

Moles of ethylene glycol = 10.8 mol/kg × 6.50kg = 70.2 mole

Now we have to calculate the mass of ethylene glycol.

Mass of ethylene glycol = Moles of ethylene glycol × Molar mass of ethylene glycol

Mass of ethylene glycol = 70.2 mole × 62.07 g/mole =4357.3 g

Now we have to calculate the volume of ethylene glycol.

$\text{Volume of ethylene glycol}=\frac{\text{Mass of ethylene glycol}}{\text{Density of ethylene glycol}}$

$\text{Volume of ethylene glycol}=\frac{4357.3g}{1.11g/mL}=3925.49mL=3.92L$

The volume of ethylene glycol is, 3.92 liters.

Now we have to calculate the boiling point of solution.

Formula used for Elevation in boiling point :

$\Delta T_b=i\times k_b\times m\\\\T_b-T^o_b=i\times k_b\times m$

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of pure water = 100^oC

k_b = boiling point constant = 0.52^oC/m

m = molality = 10.8 mol/kg

i = Van't Hoff factor = 1 (for non-electrolyte)

Now put all the given values in the above formula, we get the boiling point of a solution.

$T_b-100^oC=1\times 0.52^oC/m\times 10.8mol/kg$

T_b=105.6^oC

Therefore, the boiling point of a solution is, 105.6^oC

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