Question

The molarity of a solution prepared by dissolving 17.0 g of hydrochloric acid (HCl (aq)) in 133 mL of water is ___ M (2 decimal place).

Answer 1

The molarity is 3.50 M/L.

Step-by-step explanation:

Molarity is found to know the amount of the solute ions present in the given volume of solution. So we determine it using the ratio of the moles of solute to the volume of the solution.

As here the weight of the solute which is hydrochloric acid is given as 17g, so we need to find the moles of it. This can be done by dividing the given amount of HCl with the molecular weight of HCl.

Molecular weight of HCl = 1.01 + 35.45 = 36.46 g/mol

So the number of moles of HCl present here will be

`No.of moles = (17)/(36.46) =0.466`

So, 0.466 moles of HCl is present in the solution.

`Molarity = (No.of moles of solute)/(Volume of solution) = (0.466)/(133*10^(-3) )`

`Molarity = 3.50 M/L`

Thus, the molarity is 3.50 M/L.