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Yes your statement is true. Molar Volume is the amount of space that the molar mass(amount of matter) occupies. This is what gives us formula for density @Vasu Khera :)

$V_m=\cfrac{V}{n}$. This is a constant with a given pressure and temperature for any gases. Standard pressure and $0^{\circ}C:22,41\text{dm}^3\cdot\text{mol}^{-1}$

@Vasu Khera
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Moles is amount of matter. Molar Volume is how much volume that Mole(matter) occupies. So if a ball with mass 1 Mole occupies $1m^{3}$ volume, then the Molar Volume is $\dfrac{1m^{3}}{1 \ mole}$ or $1m^{3} \ per \ mole$

@Vasu Khera
–
A mole of something has 6.022 × 10²³ atoms, so the ball has 6.022 × 10²³ number of atoms we can say. 6.022 × 10²³ is Avogadro's Number :)

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestYes your statement is true. Molar Volume is the amount of space that the molar mass(amount of matter) occupies. This is what gives us formula for density @Vasu Khera :)

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$V_m=\cfrac{V}{n}$. This is a constant with a given pressure and temperature for any gases. Standard pressure and $0^{\circ}C:22,41\text{dm}^3\cdot\text{mol}^{-1}$

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Can you make it more clear? Please elaborate.

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What should I explain? The variables?

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No. I mean the concept. I'm pretty confused about it.

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$n=\cfrac{m}{M},m=\rho V\implies n=\cfrac{\rho V}{M}$, but $\rho=\cfrac{M}{V_m}\therefore n=\cfrac{V}{V_m}$

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$1m^{3}$ volume, then the Molar Volume is $\dfrac{1m^{3}}{1 \ mole}$ or $1m^{3} \ per \ mole$

Moles is amount of matter. Molar Volume is how much volume that Mole(matter) occupies. So if a ball with mass 1 Mole occupiesLog in to reply

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@Vasu Khera - link

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I thank you, buddy.

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No prob, chemistry is the one thing I'm good at LOL :)

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