Question

Calculate the volume in ml of a 0.555 M NaHCO3 solution that contains 14.7 g of NaHCO3. ______ ml

Answer 1

Answer:

315 mL

General Formulas and Concepts:

$Molarity=\frac{moles \hspace{3} of \hspace{3} solute}{liters \hspace{3} of \hspace{3} solution}$

Explanation:

Step 1: Define variables

0.555 M NaHCO₃

14.7 g NaHCO₃

Step 2: Define conversions

Molar Mass of Na - 22.99 g/mol

Molar Mass of H - 1.01 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of NaHCO₃ - [22.99 + 1.01 + 12.01 + 3(16.00)] g/mol = 84.01 g/mol

1 L = 1000 mL

Step 3: Find moles of solute

$14.7 \hspace{3} g \hspace{3} NaHCO_3(\frac{1 \hspace{3} mol \hspace{3} NaHCO_3}{84.01 \hspace{3} g \hspace{3} NaHCO_3} )$ = 0.174979 mol NaHCO₃

Step 4: Find amount of solution

$0.555 \hspace{3} M \hspace{3} NaHCO_3=\frac{0.174979 \hspace{3} mol \hspace{3} NaHCO_3}{x \hspace{3} L}$

$0.555 \hspace{3} M \hspace{3} NaHCO_3(x \hspace{3} L)=0.174979 \hspace{3} mol \hspace{3} NaHCO_3$

$x \hspace{3} L=\frac{0.174979 \hspace{3} mol \hspace{3} NaHCO_3}{0.555 \hspace{3} M \hspace{3} NaHCO_3}$

$x \hspace{3} L=0.315278$

Step 5: Convert

$0.315278 \hspace{3} L(\frac{1000 \hspace{3} mL}{1 \hspace{3} L} )$ = 315.278 mL

Step 6: Simplify

We are given 3 sig figs.

315.278 mL ≈ 315 mL