Question

What mass, in grams, of NaCl needs to be added to 1.6 kg of water in order to create a solution with a freezing point of -5.6 °C? The freezing point depression constant of water is 1.86 ºC/m.​

Answer 1

Step-by-step explanation:

Step-by-step explanation:

Answer:

Mass of NaCl required = 141 g

Explanation:

Given:

Mass of water = 1.6 kg = 1600 g

Freezing point of solution = -5.6 °C

Freezing point depression constant = 1.86 °C/m

To Find:

Mass of NaCl to be added

Solution:

First finding the depression in freezing point of the solution,

$\sf \Delta\:T_f=Freezing\:point\:of\:pure\:solvent-Freezing\:point\:of\:solution$

Therefore,

$\sf \Delta\:T_f=0^oC+5.6^oC$

$\sf \Delta\:T_f=5.6^oC$

The molar mass of NaCl = 23 + 35.5 = 58.5 g

Van't hoff factor for NaCl = 2

We know that,

$\boxed{\sf M_2=\dfrac{i\times K_f\times w_2\times 1000}{\Delta\:T_f\times w_1}}$

where M₂ = molar mass of the solute

$\sf K_f$ = Cryoscopic constant/Freezing point depression constant

w₂ = weight of solute

w₁ = weight of solvent

$\sf \Delta\:T_f$ = Depression in freezing point

i = van't hoff factor

Substitute the data,

$\sf 58.5=\dfrac{2\times 1.86\times w_2\times 1000}{5.6\times 1600}$

$\sf 58.5=\dfrac{3720\times w_2}{8960}$

$\sf 58.5=0.415\:w_2$

$\sf w_2=\dfrac{58.5}{0.415}$

$\sf w_2=140.96 \approx 141\:g$

Hence 141 g of NaCl must be added to the solvent.

Answer 2

Answer:

ΔTf=iKfm

Here, ΔTf= lowering in freezing point of water =7.500C

i= Van't Hoff factor =1.87

Kf=1.860C/m

m=molality

Putting the values of known variables to find m,

m=iKfΔTf=1.87×1.867.5=2.156

m= molality =mass of water in kgnumber of moles of NaCl

number of moles of NaCl=molar mass of NaClmass of NaCl=58.5mass of NaCl

m=0.065mass of NaCl/58.5

mass of NaCl=m×0.065×58.5=2.156×0.065×58.5

mass of NaCl=8.199g