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tekilochka [14]

3 years ago

15

What is the boiling point of a solution made by dissolving 23.6 g of a non-ionizing solute with a molar mass of 103.5 g/mol in 1

88 g of water?

1 answer:

vaieri [72.5K]3 years ago

7 0

<u>Answer:</u> The boiling point of solution is 100.62°C

<u>Explanation:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

Boiling point of pure water = 100°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_b$ = molal boiling point elevation constant = 0.512°C/m.g

$m_{solute}$ = Given mass of solute = 23.6 g

$M_{solute}$ = Molar mass of solute = 103.5 g/mol

$W_{solvent}$ = Mass of solvent (water) = 188 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=1\times 0.512^oC/m\times \frac{23.6\times 1000}{103.5g/mol\times 188}\\\\\text{Boiling point of solution}=100.62^oC$

Hence, the boiling point of solution is 100.62°C

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Answer:

The answer to your question is given below

Explanation:

To convert from Particles to grams, one must have a clear understanding of Avogadro's hypothesis. From the Avogadro's hypothesis, we can easily convert from mole to particles and to grams.

Avogadro's hypothesis gives us a background understanding that 1 mole of any substance contains 6.02x10^23 particles. With this in mind we can easily convert particles to grams and grams to particles.

Now consider the following example:

1. How many particles are there in 4g of carbon?

Solution:

From Avogadro's hypothesis, 1 mole of any substance contains 6.02x10^23 particles. This implies that 1 mole of carbon contains 6.02x10^23 particles.

1 mole of carbon = 12g

Now we can say that 12g of carbon contains 6.02x10^23 particles.

If 12g of carbon contains 6.02x10^23 particles,

then 4g of carbon will contain = (4x6.02x10^23) /12 = 2.01x10^23 particles.

The next example will teach us how to convert from particles to grams..

2. What mass of helium contains 4.2x10^24 particles.

Solution:

According to Avogadro's hypothesis,

1 mole of He contains 6.02x10^23 particles.

1 mole of He = 4g

We can thus, say that 4g of He contains 6.02x10^23 particles.

1f 4g of He contains 6.02x10^23 particles,

Then Xg of He contains 4.2x10^24 Particles i.e

Xg of He = (4x4.2x10^24)/(6.02x10^23)

Xg of He = 27.91g

Therefore, 27.91g of He contains 4.2x10^24 Particles

With the above illustrations I believe you have understood how to convert from grams to particles and from particles to grams.

7 0

3 years ago

Name the following ionic compounds: (a) Na2SO4, (b)Cu(NO3)2 (c) Fe2(CO3)3

Sindrei [870]

Answer and Explanation:

(a) $Na_2So_4$ : its chemical name is sodium sulfate in which sodium is present as $Na_2^+$ and sulfate is present as $So_4^{2-}$

(b) $Cu(NO_3)_2$ : Its chemical name is copper nitrate which is a inorganic compound which is mostly crystalline in nature

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4 years ago

How many moles of hcl are required to neutralize aqueous solutions of these bases:

Lemur [1.5K]

The answers are a.) 0.03 mol KOH requires 0.03 mol HCl, b.) 2 mol NH3 requires 2 mol HCl and c.) 0.1 mol Ca(OH)2 requires 0.2 mol HCl.
Solution:
We need to write the balanced equations for each reactions to find out the stoichiometry for each reactants.
a.) HCl (aq) + KOH (aq) → KCl (aq) + H2O(ℓ)
From the balanced equations, we can see that 1 HCl reacts with 1 KOH, therefore if 0.03 mol KOH is reacted then 0.03 mol HCl must also be present.

b.) HCl(aq) + NH3(aq) ) → NH4Cl(aq)
If 2 moles of NH3 are reacted then 2 moles of HCl must also be present since 1 HCl reacts with 1 NH3 from the balanced reaction.

c.) 2HCl(aq) + Ca(OH)2(s) → CaCl2(aq) + 2H2O(ℓ)
We can see that 2 HCl react with 1 Ca(OH)2, hence if 0.1 mol of Ca(OH)2 is reacted then 0.2 mol HCl must also be present.

8 0

3 years ago

Read 2 more answers

What would be the freezing point of a 1.7 mole aqueous ethylene glycol solution? The freezing point depression constant for wate

nata0808 [166]

Answer:

$-3.2^oC$

Explanation:

In order to answer this question, we need to be familiar with the law of freezing point depression. The law generally states that mixing our solvent with some particular solute would decrease the freezing point of the solvent.

This may be expressed by the following relationship:

$\Delta T_f=iK_fb$

Here:

$\Delta T_f=T_{initial}-T_{final}$ is the change in the freezing point of the solvent given its initial and final freezing point temperature values;

$i$ is the van 't Hoff factor (i = 1 for non-electrolyte solutes and i depends on the number of moles of ions released per mole of ionic salt);

$K_f$ is the freezing point depression constant for the solvent;

$b=\frac{n_{solute}}{m_{solvent}}$ is molality of the solute, defined as a ratio between the moles of solute and the mass of solvent (in kilograms).

We're assuming that you meant 1.7-molal solution, then:

$b=1.7 m$

Given ethylene glycol, an organic non-electrolyte solute:

$i=1$

The freezing point depression constant:

$K_f=1.86^oC/m$

Initial freezing point of pure water:

$T_{initial}=0.00^oC$

Rearrange the equation for the final freezing point and substitute the variables:

$T_f=T_o-iK_fb=0.00^oC-1\cdot1.86^oC/m\cdot1.7 m=-3.16^oC$

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3 years ago

Which phrase accurately describes an elliptical galaxy?

Crank

Answer:

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3 years ago