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# Find out the volume of 6.023*10²² of ammonia at stp?

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#### Explanation

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#### Explanation:

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4
Aug 17, 2017

The volume is 2.271 L.

#### Explanation:

Method 1. Using the Ideal Gas Law

We can use the Ideal Gas Law to solve this problem.

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

• $p$ is the pressure
• $V$ is the volume
• $n$ is the number of moles
• $R$ is the gas constant
• $T$ is the temperature

We can rearrange the Ideal Gas Law to get

$V = \frac{n R T}{p}$

Step 1. Calculate the moles of ammonia

n = 6.023 × 10^22 color(red)(cancel(color(black)("molecules NH"_3))) × "1 mol NH"_3/(6.022 × 10^23 color(red)(cancel(color(black)("molecules NH"_3)))) = "0.100 02 mol NH"_3

Step 2. Calculate the volume at STP

Remember that STP is defined as 0 °C and 1 bar.

$n = \text{0.100 02 mol}$
$R = \text{0.083 14 bar·L·K"^"-1""mol"^"-1}$
$T = \text{(0 + 273.15) K" = "273.15 K}$
$p = \text{1 bar}$

V = (nRT)/p = ("0.100 02" color(red)(cancel(color(black)("g"))) × "0.083 14" color(red)(cancel(color(black)("bar"))) ·"L"· color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "2.271 L"

Method 2. Using the molar volume

We know that there are 0.100 02 mol of ${\text{NH}}_{3}$.

We also know that the molar volume of a gas is 22.71 L at STP.

$V = {\text{0.100 02" color(red)(cancel(color(black)("mol NH"_3))) × "22.71 L NH"_3/(1 color(red)(cancel(color(black)("mol NH"_3)))) = "2.271 L NH}}_{3}$

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