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Anton [14]
3 years ago
5

You want to determine the concentration of a solid solute dissolved in a specified volume of liquid solvent which of the followi

ng would be the best to use?
a) volume/volume percent
b) molarity
c)mass/mass percent
d)mole fraction
Chemistry
2 answers:
Tresset [83]3 years ago
7 0

The answer is: b) molarity.

Molar concentration (molarity) is a measure of the concentration of a solute in a solution.

Unit for molarity (c = n/V) is the number of moles (n) per litre (V), having the unit symbol mol/L or sometimes mol/dm³.

For example:

m(NaOH) = 80.0 g; mass of solide solute.

n(NaOH) = m(NaOH) ÷ M(NaOH).

n(NaOH) = 80 g ÷ 40 g/mol.

n(NaOH) = 2 mol.

V(NaOH) = 500 mL ÷ 1000 mL/L.

V(NaOH) = 0.5 L; a specified volume of liquid solvent.

c(NaOH) = n(NaOH) ÷ V(NaOH).

c(NaOH) = 2 mol ÷ 0.5 L.

c(NaOH) = 4 mol/L.

Natali [406]3 years ago
6 0
Generally, chemists prefer to use morality (B) because it only invovles measuring the final volume of the solution and amount of moles of the solute

Hope this helps
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6 0
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When 25.0 ml of 0.500 m h2so4 is added to 25.0 ml of 1.00 m koh in a coffee-cup calorimeter at 23.50°c, the temperature rises to
AlladinOne [14]

The chemical reaction that occurs between the given substances is a neutralization reaction as shown below :

 

H_2SO_4 + 2KOH -> K_2SO_4 + 2H_2O

<span>1 mol            2 mol      1 mol           2 mol </span>

 

<span>The number of moles of the given substances  is calculated as shown : </span>

Number of moles of H_2SO_4 = 25.0 mL x 0.50 M = 12.5 millimoles

Number of moles of KOH = 25.0 mL x 1.00 M = 25.0 millimoles

 

As 1 mol of sulfuric acid reacts with 2 mol of KOH to give 2 mol of water, 12.5 millimoles of sulfuric acid completely reacts with 25.0 millimoles of KOH to give 25.0 millimoles of water.

Total volume of the solution = 25.0 mL + 25.0 mL = 50.0 mL.

Density of water is 1 g/mL. Use this to calculate the mass of the solution.

<span>Mass of the solution – 50.0 mL x 1 g/mL = 50.0  </span>

The specific heat of water is 4.184 J/gK. The temperature of the solution is increased from 23.5 degrees Celsius to 30.17 degrees Celsius.

The amount of heat released = 4.184 J/gK x 50.0 g x (30.17C – 23.50C) 1395 J

 

The amount of heat released per one mole of water formed can be calculated as shown :

The amount of heat released for formation of mole water = 1395 J / (25.0 m mol x 1mol/1000 m mol)

= 55,800 J

<span> </span>

8 0
3 years ago
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