Calculate the concentration of the solution produced when a 3.2g of sodium chloride is

Chemistry

1 answer:

sergejj [24]3 years ago

5 0

Answer:

The concentration of the 3.2 g of sodium chloride in 8 dm³ of water is 6.845 × 10⁻³ M

Explanation:

The concentration of a solution is given as either the ratio of mass to volume, or the ratio of volume to volume or as molarity

$Concentration, C = \dfrac{Mass}{Volume} = \dfrac{Volume}{Volume} = \dfrac{Number \ of \ moles }{Liters \ of \ solution}$

The number of moles present in 3.2 g of sodium chloride, 'n', is given as follows;

$n = \dfrac{Mass}{Molar \ mass}$

The molar mass of sodium chloride, NaCl = 58.44 g/mol

n = 3.2 g/58.44 g/mol = 0.0547570157 moles

8 dm³ = 8 Liters

Therefore, the concentration, C = 0.0547570157 moles/(8 liters) = 6.845 × 10⁻³ M

The concentration of the 3.2 g of sodium chloride in 8 dm³ of water = 6.845 × 10⁻³ Molar solution.

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Sr is an alkaline earth metal which has 2 valence electrons that when both are lost it has an oxidation state of +2. Since we know the hydroxide ion has a charge of -1, $-_{OH}$ , Then two $-_{OH}$ are needed to form compound, hence we have $SrOH_2$

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The oxidation state of zirconium is +1 since it is stated that we have zirconium(I). Converserly, perchlorate has a -1 charge which further proves that zirconium has a +1 oxidation state

The product would have a strontium diperchlorate because we know from earlier that Sr has a +2 oxidation state, and we would also have ZrOH because Zr has a +1 oxidation state and $-_{OH}$ has a -1 charge

The unbalanced equation with all chemicals formulas would be

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