# What mass of glucose, C_6H_12O_6, would be required to prepare 5 * 10^3 L of a 0.215 M solution?

Feb 29, 2016

193661.25g ${C}_{6} {H}_{12} {O}_{6}$ (before sig figs)

#### Explanation:

Molarity is defined as $M$= $\frac{m o l s}{1 L s o l u t i o n}$

from this, we can deduct that .215$M$ ${C}_{6} {H}_{12} {O}_{6}$= $\frac{.215 m o l}{1 L s o l u t i o n}$

$C$ = 12.01 * 6 = 72.06g
$H$ = 1.0079 * 12 = 12.0948g
$O$ = 16.00 * 6 = 96.00g
${C}_{6} {H}_{12} {O}_{6}$= 180.15g

$5 \cdot {10}^{3}$$L$ ${C}_{6} {H}_{12} {O}_{6}$ $\frac{.215 m o l {C}_{6} {H}_{12} {O}_{6}}{1 L s o l u t i o n}$ $\frac{180.15 g}{1 m o l {C}_{6} {H}_{12} {O}_{6}}$= 193661.25g ${C}_{6} {H}_{12} {O}_{6}$ (before sig figs)