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3 years ago

11

The rate of dissolution is affected by all of the following except the A. agitation of solute within the solvent. B. surface are

a of the solute. C. temperature of the solute. D. amount of the solute dissolved.

1 answer:

laila [671]3 years ago

8 0

Answer:

B

Explanation:

cause surface area of solute is not necessary for solution

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The concentration of copper(II) sulfate in one brand of soluble plant fertilizer is 0.070% by weight. If a 21.5 g sample of this

Semmy [17]

Answer:

The molar concentration of $Cu^{2+}$ ions in the given amount of sample is $4.73\times 10^{-5}M$

Explanation:

Given that,

Mass of sample = 21.5 g

0.07 % (m/m) of copper (II) sulfate in plant fertilizer

This means that, in 100 g of plant fertilizer, 0.07 g of copper (II) sulfate is present

So, in 20 g of plant fertilizer

, $=\frac{0.07}{100}\times 21.5}\\=0.0151g$ of copper (II) sulfate is present.

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Mass of solute (copper (II) sulfate) = 0.0151 g

Molar mass of copper (II) sulfate = 159.6 g/mol

Volume of solution = 2.0 L

$\text{Molarity of solution}=\frac{0.0151g}{159.6g/mol\times 2.0L}\\\\\text{Molarity of solution}=4.73\times 10^{-5}M$

The chemical equation for the ionization of copper (II) sulfate follows:

$CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}$

1 mole of copper (II) sulfate produces 1 mole of copper (II) ions and sulfate ions

Molarity of copper (II) ions = $4.73\times 10^{-5}M$

Hence, the molar concentration of $Cu^{2+}$ ions in the given amount of sample is $4.73\times 10^{-5}M$

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