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3 years ago

What is the mass (in grams) of 9.27x1024 molecules of methanol(CH3OH)

2 answers:

5 0

Moles of methanol = 9.27x10^24/6.02x10^23 = 15.398 moles.

Mass of methanol = moles of methanol x molar mass of methanol
= 15.398 x 32.042
= 493.38 grams.

Hope this helps!

seropon [69]3 years ago

4 0

Answer: 493 grams

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given molecules}}{\text{Avogadro's number}}=\frac{9.27\times 10^{24}}{6.023\times 10^{23}}=15.4moles$

1 mole of $CH_3OH$ weighs = 32 g

15.4 moles of $CH_3OH$ weigh = $\frac{32}{1}\times 15.4=493g$

Thus mass of $9.27\times 10^{24}$ molecules of methanol weigh 493 grams.

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ANEK [815]

Answer:

Given equation of parabola is

and

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On differentiating both sides of equation (i), we get

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Explanation:

We are given:

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To calculate the percentage composition of an element in a compound, we use the equation:

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For carbon element:

Putting values in equation 1, we get:

$\% \text{ composition of carbon element}=\frac{96}{151}\times 100\\\\\% \text{ composition of carbon element}=63.57\%$

For hydrogen element:

Putting values in equation 1, we get:

$\% \text{ composition of hydrogen element}=\frac{9}{151}\times 100\\\\\% \text{ composition of hydrogen element}=5.96\%$

For nitrogen element:

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$\% \text{ composition of nitrogen element}=\frac{14}{151}\times 100\\\\\% \text{ composition of nitrogen element}=9.28\%$

For oxygen element:

Putting values in equation 1, we get:

$\% \text{ composition of oxygen element}=\frac{32}{151}\times 100\\\\\% \text{ composition of oxygen element}=21.19\%$

Hence, the mass percentage of carbon element is 63.57 %, hydrogen element is 5.96 %, nitrogen element is 9.28 % and oxygen element is 21.19 %

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