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Eva8 [605]
3 years ago
13

What is the equation for concentration with Solutes and Solvents?​

Chemistry
1 answer:
QveST [7]3 years ago
5 0

Answer:

 Molarity = no. of moles of solute / Liter of solution

  Molality = no. of moles of solute / Kilogram of solvent

Explanation:

Solvent: Solvent is a liquid that dissolve the solute. in most cases it is water but some time other liquids gases can also be a solvent.solvent have the ability to dissolve solute and in larger volume than solute.

Solute:

It is the substance that is in less amount than solvent and dissolve in it

Solution:

When solvent and solute combines, the solute get dissolved in it and it form a solution.

Concentration:

It is to represent or describe the property of solution and refer to the amount of one constituent per total mixture or solvent.

There are two terms to represent the concentration in terms of solute and solvent.

Molarity:

It the amount of solute in moles dissolved per liter of solution. it can be represented by the following equation

         Molarity = no. of moles of solute / Liter of solution

Molality:

It is the amount of solute in moles dissolved per kilogram of solvent. it can be representd by the following equation

           Molality = no. of moles of solute / Kilogram of solvent

           

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2 years ago
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A 0.590 gram sample of a metal, M, reacts completely with sulfuric acid according to:M(s) +H2SO4(aq) --&gt; MSO4(aq) +H2(g)A vol
photoshop1234 [79]

Answer:

MM = 58.41 g

Explanation:

First, the data we have is according to the hydrogen which is exerting pressure. To solve this, we need to use the ideal gas equation:

PV = nRT (1)

the molar mass of any compound is calculated like this:

MM = m/n (2)

So, from (1) we solve for the moles (n) and then, this value is replace in (2).

However, before we do all that, we need to gather all the correct data.

All the species in the reaction are solid or aqueous state, with the exception of hydrogen, which is gaseous. Hydrogen is collected over water, therefore, is exerting some pressure too. The problem is not indicating if the acid or any other species is exerting pressure, so we will assume that only hydrogen and water are exerting pressure.

The total pressure exerted by the system would be:

P = Pw + PH2 (3)

We already know the total pressure which is 756 torr.

This experiment is taking place at 25 °C (298.15 K), and at this temperature, we have a reported value for water pressure which is 23.8 Torr.

Let's solve for PH2:

PH2 = P - Pw

PH2 = 756 - 23 = 733 Torr

Now, with this value, and the volume and temperature, we can calculate the moles of H2:

n = PV/RT

But first, let's convert the pressure to atm:

PH2 = 733 Torr / 760 torr * 1 atm = 0.9644 atm

now, solving for n:

n = 0.9644 * (0.255) / 0.082 * 298.15

n = 0.0101 moles

Now that we have the moles, we know that the metal and the hydrogen has a mole ratio of 1:1 according to the reaction, so, this means that:

moles M = moles H2 = 0.0101 moles

We have the moles of the metal and the mass, we can calculate the molar mass using expression (2):

MM = 0.590/0.0101

MM = 58.41 g/mol

This is the molar mass of the metal

8 0
3 years ago
If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.
kicyunya [14]
Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

CH_{3}COOH \ \textless \ ---\ \textgreater \   H^{+} + CH_{3}COO^{-}

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

pH = 2.88 ==> [H^{+}]  = 10^{-2.88} =  0.001 moldm^{-3}

The change in Concentration Δ [CH_{3}COOH]= 0.001 moldm^{-3}


                                  CH3COOH          H+           CH3COOH    
Initial  moldm^{-3}                      x           0                     0
                                                                                                                       
Change moldm^{-3}        -0.001            +0.001           +0.001
                                                                                                       
Equilibrium moldm^{-3}      x- 0.001      0.001             0.001
                                                                              

Since the k_{a} value is so small, the assumption 
[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium} can be made.

k_{a} = [tex]= 1.8*10^{-5}  =  \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} =  \frac{0.001^{2}}{x}

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

         2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps! 



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Answer:

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Explanation:

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