Question

How many moles and grams of $HCl$ are present in $300.00$ ml of $12.0$ M solution?

Answer 1

$\text{Number of moles = molarity}\times \text{volume(litre)}$

$\therefore$ Number of $moles=12\times 0.300=3.6moles$

$3.6moles$ in grams will be :

$3.6\times 36.5g=131.4g$ of $HCl$.