What mass of glucose, $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ , would be required to prepare $5 \cdot 10^{3}$ $5 * 10^3$ $L$ $L$ of a $0.215$ $0.215$ $M$ $M$ solution?

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What mass of glucose, $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ , would be required to prepare $5 \cdot 10^{3}$ $5 * 10^3$ $L$ $L$ of a $0.215$ $0.215$ $M$ $M$ solution?

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Answer 1 · 2016-02-29T13:25:41

193661.25g $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ (before sig figs)

Explanation:

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

from this, we can deduct that .215 $M$ $M$ $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ = $\frac{.215 m o l}{1 L s o l u t i o n}$ $(.215mol)/(1L solution)$

$C$ $C$ = 12.01 * 6 = 72.06g
$H$ $H$ = 1.0079 * 12 = 12.0948g
$O$ $O$ = 16.00 * 6 = 96.00g
$C_{6} H_{12} O_{6}$ $C_6H_12O_6$ = 180.15g

Start with what is given

$5 \cdot 10^{3}$ $5*10^3$ $L$ $L$ $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ $\frac{.215 m o l C_{6} H_{12} O_{6}}{1 L s o l u t i o n}$ $(.215 molC_6H_12O_6)/(1L solution)$ $\frac{180.15 g}{1 m o l C_{6} H_{12} O_{6}}$ $(180.15g)/(1mol C_6H_12O_6)$ = 193661.25g $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ (before sig figs)

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What mass of glucose, $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ , would be required to prepare $5 \cdot 10^{3}$ $5 * 10^3$ $L$ $L$ of a $0.215$ $0.215$ $M$ $M$ solution?

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What mass of glucose, C6H12O6C_6H_12O_6, would be required to prepare 5⋅1035 * 10^3 LL of a 0.2150.215 MM solution?

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What mass of glucose, $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ , would be required to prepare $5 \cdot 10^{3}$ $5 * 10^3$ $L$ $L$ of a $0.215$ $0.215$ $M$ $M$ solution?