One.liter of an unknown gas weigh 24.3 grams at.standard pressure and -273 degree centigrade one possible formula of gas is 1 CO2 , 2 H2 3, O2 4, SO2?

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One.liter of an unknown gas weigh 24.3 grams at.standard pressure and -273 degree centigrade one possible formula of gas is 1 CO2 , 2 H2 3, O2 4, SO2?

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Answer 1 · 2018-07-05T00:03:27

None of the above! Your "gas" is a solid.

But let's treat this problem in a theoretical manner anyway...

Here we just seek the molar mass of the gas, and hence find its identity... The assumption is that it is ideal at high temperature and low pressure, neither of which is true here... but it is implied in the question to do so anyway

$P V = n R T$ $PV = nRT$

$P$ $P$ is pressure in $bar$ $"bar"$ .

$V$ $V$ is volume in $L$ $"L"$ .

$n$ $n$ is mols of ideal gas.

$R = 0.083145 L \cdot bar/mol \cdot K$ $R = "0.083145 L"cdot"bar/mol"cdot"K"$ is the universal gas constant.

$T$ $T$ is temperature in $K$ $"K"$ .

Now, we can solve for the mols, because we are given the mass.

$n = \frac{P V}{R T}$ $n = (PV)/(RT)$

Being modern people here, we use the post-1982 IUPAC ruling of $1 bar$ $"1 bar"$ standard pressure, and assume you mean $- 272^{\circ} C$ $-272^@ "C"$ to get the unbelievable $1.15 K$ $"1.15 K"$ :

$n = \frac{1 bar \cdot 1 L}{0.083145 L \cdot bar/mol \cdot K \cdot 1.15 K}$ $n = ("1 bar" cdot "1 L")/("0.083145 L"cdot"bar/mol"cdot"K" cdot "1.15 K")$

$=$ $=$ $10.5 mols$ $"10.5 mols"$ ...

...for which we apparently get:

$M_{m} = \frac{24.3 g}{10.5 mols} = 2.32 g/mol$ $color(red)(M_m) = "24.3 g"/"10.5 mols" = color(red)("2.32 g/mol")$

The only one remotely close is $H_{2}$ $"H"_2$ , with $M_{m} = 2.016 g/mol$ $M_m = "2.016 g/mol"$ . It's red because it does not represent physical reality. Examine the phase diagram:

![https://www.researchgate.net/]( useruploads.socratic.org )

And we see that at $1.15 K$ $"1.15 K"$ and $1 bar$ $"1 bar"$ , hydrogen gas is... a SOLID!

Answer 2 · 2018-07-05T00:39:46

.See below

Explanation:

Alternative method:
Given this information, we need to find the molar mass of this gas in grams/mol to truly determine it's identity.
We will use the gas law constant: $\frac{0.0821 L \cdot atm}{mol \cdot K}$ $(0.0821 " L"*"atm")/("mol"*K)$

We need the temperature in Kelvin so:
$C + 273.15 = K = - 272^{\circ} C + 273.15 = 1.15 K$ $C+273.15=K=-272^@C+273.15= 1.15K$

Use dimensional analysis until you're left with "g/mol":
$24.3" g Gas"xx(0.0821 cancel" L"*cancel"atm")/("mol"*cancel"K")xx(1.15cancel" K")/(1 cancel"atm")xx1/(1cancel"L")approx 2.29 g/"mol"$

This is about the molar mass of $H_2$ , so the gas must be hydrogen gas, however as the answer above stated, the conditions are unrealistic

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One.liter of an unknown gas weigh 24.3 grams at.standard pressure and -273 degree centigrade one possible formula of gas is 1 CO2 , 2 H2 3, O2 4, SO2?

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