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4 years ago

What effect does increasing the concentration of a dissolved solute have on each of the colligative properties?

Chemistry

1 answer:

Igoryamba4 years ago

7 0

Answer:

If we increase the concentration of a dissolved solute, the solution would have a vapor pressure so much low, the boiling temperature for the solution will be so high, freezing point for the solution will be so much low and the osmotic pressure will be higher.

Colligative properties always depends on dissolved particles (solute)

Explanation:

These are the colligative properties

- Vapor pressure lowering

ΔP = P° . Xm

Vapor pressure of pure solvent - Vapor pressure of solution.

If we add more solute, it would raise the Xm, so the solution would have a vapor pressure so much low.

Vapor pressure pure solvent - Vapor pressure solution ↑ = P° . Xm ↑

- Boiling point elevation

ΔT = Kb . m

When we add more solute, we are increasing the molality.

↑T° boiling of solution - T° boiling pure solvent = Kf . m ↑

Boiling temperature for the solution will be so high.

- Freezing point depression

When we add more solute, we are increasing the molality.

ΔT = Kf . m

T° fussion of pure solvent - ↓T° fussion of solution = Kf . m↑

Freezing point for the solution will be so much low.

- Osmotic pressure

π = M . R . T

When we add solute, molarity is increasing. Therefore the osmotic pressure will be higher.

π↑ = M↑ . R . T

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Can you attach the following concept map on chemical reactions?

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3 years ago

Two methanol-water mixtures are contained in separate tanks. The first mixture contains 40.0 wt% methanol and the second contain

RoseWind [281]

Answer:

$M_{per}= 52.86$

$W_{per}=47.14$

Explanation:

<u>First mixture</u>:

40 wt% methanol - 60 wt% water 200 kg

$m_{met1}=200 kg * 0.4= 80 kg$

$m_{wat1}=200 kg * 0.6= 120 kg$

<u>Second mixture</u>:

70 wt% methanol - 30 wt% water 150 kg

$m_{met2}=150 kg * 0.7= 105 kg$

$m_{wat2}=150 kg * 0.3= 45 kg$

Final mixture:

$m_{metF=80 kg + 105 kg= 185 kg$

$m_{watF}=120 kg + 45 = 165 kg$

$M_{per}=\frac{185 kg}{185 kg + 165 kg}*100= 52.86$

$W_{per}=\frac{165 kg}{185 kg + 165 kg}*100=47.14$

If, the compositions are constant, the only variables are the mass of each mixture used in the final one, so there can be only one independent balance.

8 0

3 years ago

Calcula la masa atómica del Hierro y las partículas subatómicas de cada uno de sus isótopos. Fe-54 (5.82%), Fe-56 (91.66%), Fe-5

tino4ka555 [31]

Answer:

La masa atómica del hierro es 55.847 gramos por mol.

Explanation:

Las masas molares de $Fe_{54}$ , $Fe_{56}$ , $Fe_{57}$ y $Fe_{58}$ son 53.940 gramos por mol, 55.935 gramos por mol, 56.935 gramos por mol y 57.933 gramos por mol, respectivamente. La masa atómica del hierro se determina mediante el siguiente promedio ponderado:

$M_{Fe} = \frac{5.82}{100}\times \left(53.940\,\frac{g}{mol} \right)+\frac{91.66}{100}\times (55.935\,\frac{g}{mol})+\frac{2.19}{100}\times \left(56.935\,\frac{g}{mol} \right)+\frac{0.33}{100}\times \left(57.933\,\frac{g}{mol} \right)$

$M_{Fe} = 55.847\,\frac{g}{mol}$

La masa atómica del hierro es 55.847 gramos por mol.

8 0

3 years ago

What volume of Co2 (carbon (iv) oxide)

hram777 [196]

Answer:

2.1056L or 2105.6mL

Explanation:

We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:

Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol

Mass of Na2CO3 = 10g

Mole of Na2CO3 =.?

Mole = mass /molar mass

Mole of Na2CO3 = 10/106

Mole of Na2CO3 = 0.094 mole

Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:

Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CO2.

Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.

Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:

1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.

Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L

Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL

7 0

4 years ago