Question

a chemist adds 195.0 ml of a $8.14 \times 10^{-6}$ mol/l zinc oxalate ($\text{znc}_2\text{o}_4$) solution to a reaction flask. calculate the micromoles of zinc oxalate the chemist has added to the flask. round your answer to 3 significant digits. $square$ $\mu$ mol

Explanation:

Step1: Convert volume to liters

$195.0\ \text{mL} = 195.0 \times 10^{-3}\ \text{L} = 0.1950\ \text{L}$

Step2: Calculate moles of zinc oxalate

$\text{Moles} = \text{Concentration} \times \text{Volume}$
$\text{Moles} = 8.14 \times 10^{-6}\ \text{mol/L} \times 0.1950\ \text{L} = 1.5873 \times 10^{-6}\ \text{mol}$

Step3: Convert moles to micromoles

$1\ \text{mol} = 10^6\ \mu\text{mol}$
$\text{Micromoles} = 1.5873 \times 10^{-6}\ \text{mol} \times 10^6\ \mu\text{mol/mol} = 1.5873\ \mu\text{mol}$

Step4: Round to 3 significant digits

Round $1.5873$ to 3 significant digits: $1.59$

Answer:

$1.59$ μ mol