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3 years ago

Assuming ideal behavior, what is the freezing point of a 0.030 m solution of Al2(SO4)3 ?

1 answer:

3 0

The freezing point of ideal solutions lower respect the freezing point of the pure solvent following the freezing depression formula:

ΔT = i * Kf * m

Where:

- i is the vant'Hoff, which is the number of partilces (ions in this case) that are the solute forms when dissociate.

- kf is cryoscopic constant of the pure solvent (water because we assume it is an aqueous solution).

Then kf = - 1.86 °C / mol

- and m is the molality, m = 0.030 m

i = 2 + 3 = 5, given the one molecule of Al2 (SO4)3 produces 2 ions of Al(+) and 3 ions of SO4 (2-).

Then, ΔT = 5 * 1.86°C/m * 0.030 m = 0.279 °C.

That is the depression of the freezing point which you have to subtract from the freezing point of the pure water => 0°C - 0.279°C = - 0.279 °C.

Answer: - 0.279°C

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What volume (in L) of a 1.25 M

sleet_krkn [62]

Answer:

V1= 0.305L

Explanation:

To find the initial volume of 1.25M potassium fluoride needed to make tge dilution specified in the question, we can use: C1 × V1 = C2 × V2

Since the question wants the volume in litres, convert 455 mL to L

455/ 1000

= 0.455 L

Now make the substitution

1.25 × V1 = 0.838 × 0.455

Rearrange to make V1 the subject

V1=

$\frac{0.838 \times 0.455}{1.25 } = 0.305$

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3 years ago

How many moles of sodium hydroxide would react with 1 Mole of sulphuric acid?

Paul [167]

Answer:

Two moles.

Explanation:

Sulphuric (sulfuric) acid $\rm H_2SO_4$ is a diprotic acid. When one mole of $\rm H_2SO_4$ molecules dissolve in water, two moles of $\rm H^{+}$ ions would be produced.

$\rm H_2SO_4 \to 2\, H^{+} + {SO_4}^{2-}$ .

On the other hand, sodium hydroxide $\rm NaOH$ is a monoprotic base. When one mole of $\rm NaOH$ formula units dissolve in water, only one mole of hydroxide ions $\rm OH^{-}$ would be produced.

$\rm NaOH \to Na^{+} + OH^{-}$ .

Note that $\rm H^{+}$ and $\rm OH^{-}$ react at a one-to-one ratio:

$\rm H^{+} + OH^{-} \to H_2O$ .

As a result, it would take $2\; \rm mol$ of $\rm OH^{-}$ to react with the $\rm 2\; mol$ of $\rm H^{+}$ that was released when $1\; \rm mol$ of $\rm H_2SO_4$ is dissolved in water. Since one mole of $\rm NaOH$ formula units could produce only one mole of $\rm OH^{-}$ , it would take $\rm 2\; mol$ of $\rm NaOH$ formula units to produce that $2\; \rm mol$ of $\rm OH^{-}$ for reacting with $1\; \rm mol$ of $\rm H_2SO_4$ .

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3 years ago

HELP ASAP PLEASE!!!

sveticcg [70]

1. Pure substances cannot be separated into any other kinds of matter, while a mixture is a combination of two or more pure substances.

2. A pure substance has constant physical and chemical properties, while mixtures have varying physical and chemical properties (i.e., boiling point and melting point).

3. A pure substance is pure, while a mixture is impure.

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2 years ago

How much calcium hypochlorite (65% strength) is needed to make 200 L of 2% hypochlorite solution? (Assume that a 1% solution is

g100num [7]

To solve this we use the equation,

M1V1 = M2V2

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3 years ago

Mass=4grams Density=2g/ml

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Answer:

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