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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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A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62
vesna_86 [32]

Answer: The concentration of H_2SO_4 is 0.234 M

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity H_2SO_4 = 2

M_1 = molarity of H_2SO_4 solution = ?

V_1 = volume of  H_2SO_4 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 0.375 M

V_1 = volume of  NaOH solution = 62.5 ml

Putting in the values we get:

2\times M_1\times 50.0=1\times 0.375\times 62.5

M_1=0.234M

Therefore concentration of H_2SO_4 is 0.234 M

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