Question

Calculate the volume at NTP occupied by 6.25 g of nitrogen.

Answer 1

At $NTP$ , $P=1atm$ , $T=293K$

$\therefore V=\frac{nRT}{P}=\frac{\frac{6.25}{28}\times 0.0821\times 293}{1}$

$\therefore V=5.369L$

$\therefore Volume=5.37L$