Question

A chemist adds 170ml of 1.99 x 10-5 mmol/L a mercury(I) chloride solution to a reaction flask. Calculate the micromoles of mercury(I) chloride the chemist has added to the flask. Round your answer to significant digits.

Answer 1

Answer: $3.38\times 10^{-9}$ micromoles of mercury(I) chloride

Explanation:

Molarity is defined as the number of moles of solute dissolved per Liter of the solution.

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$     .....(1)

Molarity of mercury(I) chloride solution = $1.99\times 10^{-5}mmol/L$

Volume of solution = 170 mL

Putting values in equation 1, we get:

$1.99\times 10^{-5}mmol/L=\frac{\text{milli moles of mercury(I) chloride }\times 1000}{170ml}\\\\\text{milli moles of mercury(I) chloride}=\frac{1.99\times 10^{-5}mmol/L\times 170}{1000}=0.338\times 10^{-5}mmol$

$1mmol=0.001\mu mol$

$0.338\times 10^{-5}mmol=3.38\times 10^{-9}\mu mol$

Thus $3.38\times 10^{-9}$ micromoles of mercury(I) chloride the chemist has added to the flask.