Question

Calculate the mass of $50\,cc$ of $CO$ at S.T.P. $[C = 12,\,O = 16]$

Answer 1

Hint: $1$ mole of a gas at S.T.P occupies a volume of $22.4$ litres and possesses mass in grams equal to its molecular mass. $1$ cubic centimetre$ = {10^{ - 3}}$ litre.
Formula used:
 $1$ mole$ = $gram molecular mass $ = 22.4$ litres of gas at STP
Gram molecular mass $ = $sum of atomic masses
$1$ cubic centimetre$ = {10^{ - 3}}$ litre

Complete step by step answer:
The atomic masses of C and O are given $[C = 12,\,O = 16]$.
The gram molecular mass of carbon monoxide ($CO$)$ = 12 + 16 = 28\,g$
In gases, a mole is defined as that amount of the gas which has a volume of $22.4$ litres at S.T.P.
Therefore,$1$ mole$ = $gram molecular mass $ = 22.4$ litres of gas at STP
Hence, $1$ mole of CO is $28\,g$ and it occupies $22.4$ litre volume.
 \[
   \Rightarrow \,22.4\,litres\, = 28\,g\,CO \\
   \Rightarrow \,1\,litre\, = \,\dfrac{{28}}{{22.4}}g\,CO \\
   \Rightarrow \,1\,litre\, = \,1.25\,g\,CO \\
  \]
Now,
 $
  1\,cc\, = \,{10^{ - 3}}\,litre \\
  \therefore \,50\,cc = 50 \times {10^{ - 3}}litre \\
   \Rightarrow 50\,cc = 0.05\,litre \\
  $
Therefore,
 $
  0.05\,litre = \,1.25 \times 0.05\,g\,CO \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,0.0625\,g\,CO \\
  $

Hence, $50\,cc$ of $CO$ at S.T.P is $0.0625$g.

Note:
Here, $1\,cc$ (cubic centimetre) of the given mass was converted to litres but we can also convert the litres of gas at STP to $cc$ $(1litre = 1000cc)$. According to Avogadro’s hypothesis ‘Equal volumes of different gases under similar conditions of temperature and pressure contain an equal number of molecules’. This means that $6.022 \times {10^{23}}$ molecules of any gas at STP (i.e., standard temperature and pressure, ${0^ \circ }C$ and atmospheric pressure) must have the same volume. This volume has been experimentally found to be $22.4$ litres at STP (${0^ \circ }C$, $1$atm or $1.01$ bar pressure) and is called molar volume.