Math, asked by dinubhai79, 2 months ago

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.​

Answers

Answered by sairam1919
30

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.

Answered by Aloneboi26
1

Answer:

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.

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