Question

a chemist adds 200.0 ml of a 0.050 mmol/l sodium nitrate (nano₃) solution to a reaction flask. calculate the mass in milligrams of sodium nitrate the chemist has added to the flask. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Calculate the amount of substance in millimoles

Use the formula $n = cV$, where $c$ is the concentration and $V$ is the volume. Given $c = 0.050\ mmol/L$ and $V=200.0\ mL = 0.2000\ L$.
$n = 0.050\ mmol/L\times0.2000\ L= 0.010\ mmol$

Step2: Calculate the molar mass of $NaNO_3$

The molar mass of $Na$ is approximately $23\ g/mol$, $N$ is approximately $14\ g/mol$, and $O$ is approximately $16\ g/mol$. So the molar mass of $NaNO_3$ is $M = 23 + 14+3\times16=85\ g/mol = 85\ mg/mmol$.

Step3: Calculate the mass in milligrams

Use the formula $m=nM$. Substitute $n = 0.010\ mmol$ and $M = 85\ mg/mmol$ into it.
$m=0.010\ mmol\times85\ mg/mmol = 0.85\ mg$

Answer:

$0.85\ mg$