# What is the molarity of 0.080 moles NaHCO_3 in 1250 mL of solution?

$\text{Molarity = 0.064} \cdot m o l \cdot {L}^{-} 1$
$\text{Molarity}$ $=$ $\text{Amount of substance (moles)"/"Volume of solution (L)}$
$=$ $\frac{0.080 \cdot m o l}{1.250 \cdot L}$ $=$ 0.064"*mol*L^-1 with respect to sodium bicarbonate.