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Sophie [7]
3 years ago
7

A chemist prepares a solution of mercury(II) iodide HgI2 by measuring out 0.0067μmol of mercury(II) iodide into a 350.mL volumet

ric flask and filling the flask to the mark with water. Calculate the concentration in /μmolL of the chemist's mercury(II) iodide solution. Round your answer to 2 significant digits.
Chemistry
1 answer:
yaroslaw [1]3 years ago
8 0

Answer:

\large\boxed{0.019\, \mu \text{mol/L}}

Explanation:

\text{Concentration } = \dfrac{\text{moles}}{\text{litres}}\\\\\text{c} = \dfrac{n}{V}

1. Convert millilitres to litres  

\text{V = 350. mL} \times \dfrac{\text{1 L}}{\text{1000 mL}} = \text{0.3500 L}

2. Calculate the concentration

c = \dfrac{0.0067\, \mu\text{mol}}{\text{0.3500 L}} = \mathbf{0.019 \,  \mu}\textbf{mol/L}\\\\\text{The concentration of mercury(II) iodide is $\large\boxed{\mathbf{0.019\, \mu} \textbf{mol/L}}$}.

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So, n = n'

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<h3>Concentration of ammonia solution</h3>

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