Question

how many moles of alcohol (C2H5OH) are there in 4.88 x 10^4 L? The density of alcohol is 0.79 g/mL.

Answer 1

Answer on Question #39579 - Chemistry - Other

Question

How many moles of alcohol (C2H5OH) are there in $4.88 \times 10^4$ L? The density of alcohol is $0.79\ \mathrm{g/mL}$.

Answer:

Mass of $\mathrm{C_2H_5OH}$ equals:

$$m = \rho V$$

$\rho$ – Density of alcohol, $\rho = 0.79\ \mathrm{g/mL} = 790\ \mathrm{g/l}$.

$V$ – Volume of alcohol, $V = 4.88 \times 10^4$ L:

$$m = 790 \cdot 4.88 \cdot 10^4 = 3.8552 \cdot 10^7\ \mathrm{g}$$

Number of moles equals:

$$n = \frac{m}{M}$$

$m$ – Mass of $\mathrm{C_2H_5OH}$, g.

$M$ – Molar mass of $\mathrm{C_2H_5OH}$, equals:

$$M = 2M(C) + 6M(H) + M(O) = 2.12 + 6.1 + 16 = 46\ \frac{\mathrm{g}}{\mathrm{mole}}$$

Then number of moles in $10.0\ \mathrm{g}$ of $\mathrm{C_2H_5OH}$ equals:

$$n = \frac{3.8552 \cdot 10^7}{46} = 838087\ \mathrm{moles} = 8.4 \cdot 10^5\ \mathrm{moles}$$

Answer: $n = 8.4 \cdot 10^5$ moles.