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Nat2105 [25]
3 years ago
11

Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is

1.86°c/m. remember to include the value of i.
Chemistry
2 answers:
Citrus2011 [14]3 years ago
8 0
Depression in freezing point (ΔT_{f}) = K_{f}×m×i,
where, K_{f} = cryoscopic constant = 1.86^{0} C/m,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For NaNO_{3})

Thus, (ΔT_{f}) = 1.86 X 0.0085 X 2 = 0.03162^{0}C

Now, (ΔT_{f}) = T^{0} - T
Here, T = freezing point of solution
T^{0} = freezing point of solvent = 0^{0}C
Thus, T = T^{0} - (ΔT_{f}) = -0.03162^{0}C
SashulF [63]3 years ago
4 0

Answer : The freezing point of a solution is, 0.32^oC

Explanation :

First we have to calculate the Van't Hoff factor (i) for NaNO_3.

The dissociation of NaNO_3 will be,

NaNO_3\rightarrow Na^++NO_3^-

So, Van't Hoff factor = Number of solute particles = Na^++NO_3^- = 1 + 1 = 2

Now we have to calculate the freezing point of a solution.

Formula used for lowering in freezing point :

\Delta T_f=i\times k_f\times m

or,

T_f^o-T_f=i\times k_f\times m

where,

\Delta T_f = change in freezing point

T_f = temperature of solution = ?

T^o_f = temperature of pure water = 0^oC

k_f = freezing point constant  = 1.86^oC/m

m = molality  = 0.08500 m

i = Van't Hoff factor = 2

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

0^oC-T_f=2\times (1.86^oC/m)\times 0.08500m

T_f=0.32^oC

Therefore, the freezing point of a solution is, 0.32^oC

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Se tiene una solución acuosa 2M de carbonato de potasio. Expresar su concentración en %p/v y Normalidad.
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Answer:

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Explanation:

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Masa K2CO3 -Masa molar: 138.205g/mol-

2moles * (138.205g/mol) = 276g K2CO3

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1L * (1000mL/1L) = 1000mL

%p/V:

276g K2CO3 / 1000mL * 100

<h3>%p/V = 27.6%</h3>
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658 mL of 0.250 M HCl solution is mixed with 325 mL of 0.600 M HCl solution. What is the molarity of the resulting solution
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The molarity of the resulting solution obtained by mixing 658 mL of 0.250 M HCl solution with 325 mL of 0.600 M HCl solution is 0.366 M

We'll begin by calculating the number of mole of HCl in each solution. This can be obtained as follow:

<h3>For solution 1:</h3>

Volume = 658 mL = 658 / 1000 = 0.658 L

Molarity = 0.250 M

<h3>Mole of HCl =?</h3>

Mole = Molarity x Volume

Mole of HCl = 0.250 × 0.658

<h3>Mole of HCl = 0.1645 mole</h3>

<h3>For solution 2:</h3>

Volume = 325 mL = 325 / 1000 = 0.325 L

Molarity = 0.6 M

<h3>Mole of HCl =?</h3>

Mole = Molarity x Volume

Mole of HCl = 0.6 × 0.325

<h3>Mole of HCl = 0.195 mole</h3>

  • Next, we shall determine the total mole of HCl in the final solution. This can be obtained as follow:

Mole of HCl in solution 1 = 0.1645 mole

Mole of HCl in solution 2 = 0.195 mole

Total mole = 0.1645 + 0.195

<h3>Total mole = 0.3595 mole</h3>

  • Next, we shall determine the total volume of the final solution.

Volume of solution 1 = 0.658 L

Volume of solution 2 = 0.325 L

Total Volume = 0.658 + 0.325

<h3>Total Volume = 0.983 L</h3>

  • Finally, we shall determine the molarity of the resulting solution.

Total mole = 0.3595 mole

Total Volume = 0.983 L

<h3>Molarity =?</h3>

Molarity = mole / Volume

Molarity = 0.3595 / 0.983

<h3>Molarity = 0.366 M</h3>

Therefore, the molarity of the resulting solution is 0.366 M

Learn more: brainly.com/question/25342554

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