Question

missed this? watch kcv: solution concentration, we: using molarity in calculations; read section 5.2. you can click on the review link to access the section in your e - text. a chemist wants to make 8.00 l of a 0.370 m nano₃ solution. part a what mass of nano₃ (in g) should the chemist use? express the mass in grams to three significant figures. view available hint(s)

Explanation:

Step1: Calculate moles of NaNO₃

Use the formula $n = M\times V$, where $M$ is molarity and $V$ is volume. Given $M = 0.370\ M$ and $V=8.00\ L$.
$n = 0.370\ mol/L\times8.00\ L=2.96\ mol$

Step2: Calculate molar mass of NaNO₃

The molar mass of Na (sodium) is approximately $22.99\ g/mol$, N (nitrogen) is approximately $14.01\ g/mol$, and O (oxygen) is approximately $16.00\ g/mol$. For NaNO₃, $M_{r}=22.99 + 14.01+3\times16.00=85.00\ g/mol$

Step3: Calculate mass of NaNO₃

Use the formula $m = n\times M_{r}$. Substitute $n = 2.96\ mol$ and $M_{r}=85.00\ g/mol$.
$m=2.96\ mol\times85.00\ g/mol = 251.6\ g$
Rounding to three - significant figures, $m = 252\ g$

Answer:

$252\ g$