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a solution is prepared by dissolving 55g of CaCl2 into 300g of water. if the Kf of water is -1.86 c/m, shat is the freezing poin

Zepler [3.9K]

Answer:

Relation between , molality and temperature is as follows.

T =

It is also known as depression between freezing point where, i is the Van't Hoff factor.

Let us assume that there is 100% dissociation. Hence, the value of i for these given species will be as follows.

i for = 3

i for glucose = 1

i for NaCl = 2

Depression in freezing point will have a negative sign. Therefore, d

depression in freezing point for the given species is as follows.

=

Therefore, we can conclude that given species are arranged according to their freezing point depression with the least depression first as follows.

Glucose < NaCl <

Explanation:

3 0

3 years ago

What is the largest contributor to the greenhouse effect?

Sloan [31]

Answer:

burning of fossil fuels

Explanation:

7 0

3 years ago

First, get about 20 mL of the quinine stock solution in a clean beaker to prepare 250 mL of about 2 ppm quinine in 0.05 M H2SO4.

iris [78.8K]

Answer:

Volume of acid, Va=250mL; Volume of quinine,Vb=20mL; Molarity of acid, Ma=0.05M.

Molar mass of acid= H2+S+O4= 2+32+(16X4)= 2+32+64=98g

Concentration of acid, Ca= Molar mass of acid/ Ma =98/0.05=1960g/mol

Explanation: To calculate concentration of quinine, Cb is as follow

Va*Ca=Vb*Cb

∴ Cb=Va*Ca/Vb =250*1960/20 =24500g/mol

5 0

3 years ago

Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depr

lutik1710 [3]

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C

Using the equation,

Δ $T_{f}$ = i $K_{f}$ m

where:

Δ $T_{f}$ = change in freezing point (unknown)

i = Van't Hoff factor

$K_{f}$ = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = $\frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }$

molal concentration = 0.1219m

Now, putting in the values to the equtaion Δ $T_{f}$ = i $K_{f}$ m we get,

Δ $T_{f}$ = 2 × 1.86 × 0.1219

Δ $T_{f}$ = 0.4536°C

So, Δ $T_{f}$ of solution is,

Δ $T_{f_{solution} }$ = 0.00°C - 0.45°C

Δ $T_{f_{solution} }$ = - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C

Learn more about freezing point here;

brainly.com/question/3121416

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7 0

1 year ago

Which segement of the heating curve shown above represents an increase in the potential energy, but no change in the kinetic ene

Ede4ka [16]

Answer: Option (b) is the correct answer.

Explanation:

Kinetic energy is defined as the energy obtained by the molecules of an object due to their motion.

Also, it is known that kinetic energy is directly proportional to temperature.

Mathematically, K.E = $\frac{3}{2}kT$

where, T = temperature

Whereas potential energy is defined as the energy obtained by an object due to its position.

Mathematically, P.E = mgh

where, m = mass

g = acceleration due to gravity

h = height

Therefore, in the given curve when temperature remains constant then kinetic energy of molecules will also remain.

Hence, we can conclude that the segment QR represents an increase in the potential energy, but no change in the kinetic energy.

7 0

2 years ago