Question

The number of molecules of methane ore present in $5.6$ litres of methane at NTP is $x\times {10}^{23}$. What is the value of $x$?

Answer 1

atomic mass & molecule mass

$V=5.6l$

$\frac{N}{6\times {10}^{23}}=\frac{5.6l}{22.4l}$

$\frac{x\times {10}^{23}}{6\times {10}^{23}}=\frac{5.6}{22.4}$

$x=1.5$