Question

homework 5 exercise 5.33 - enhanced - with expanded hints 8 of 14 missing this? watch kcv solution concentration, iwe: 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 6.00 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) incorrect; try again beginning with the given volume, use the concentration (in units of molarity) as a starting point to convert to the number of moles. then, consult the periodic table for the relationship between moles and mass. you may want to review.

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 $V=6.00\ L$. So $n = 0.370\ mol/L\times6.00\ L= 2.22\ mol$.

Step2: Calculate molar - mass of NaNO₃

The molar - mass of NaNO₃: $M_{NaNO₃}=22.99\ g/mol + 14.01\ g/mol+3\times16.00\ g/mol=84.99\ g/mol$.

Step3: Calculate mass of NaNO₃

Use the formula $m = n\times M$. Substitute $n = 2.22\ mol$ and $M = 84.99\ g/mol$ into the formula. So $m=2.22\ mol\times84.99\ g/mol\approx189\ g$.

Answer:

189 g