# Question a1dd4

$= 2.4 m o l$
Strength of $N a C l$Solution =0.8%M=0.8/100(mol)/L#
So the amount of $N a C l$present in 300L of this solution is
$= \frac{0.8}{100} \frac{m o l}{\cancel{L}} \times 300 \cancel{L} = 2.4 m o l$