Question

Part IV. For each trial, calculate the number moles of 6.0 M HCl used in the reaction. Report your answer using 4 digits. Note that it is 1 or 2 digits beyond the number of significant figures, i.e. 0.2345 moles.

Answer 1

This is an incomplete question, here is a complete question.

For each trial, calculate the number moles of 6.0 M HCl used in the reaction?

Trial 1 : Volume of HCl = 15.0ml

Trial 2 : Volume of HCl = 14.9ml

Trial 3 : Volume of HCl = 15.2ml

Answer :

The number moles of HCl for trial 1, 2 and 3 is, 0.090 mol, 0.089 mol and 0.091 mol

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

`\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}`

In this question, the solute is HCl.

Now we have to calculate the number of moles of HCl for trial 1.

Volume of HCl = 15.0 mL = 0.015 L

`6.0M=\frac{\text{Moles of HCl}}{0.015L}`

`\text{Moles of HCl}=0.090mol`

Now we have to calculate the number of moles of HCl for trial 2.

Volume of HCl = 14.9 mL = 0.0149 L

`6.0M=\frac{\text{Moles of HCl}}{0.0149L}`

`\text{Moles of HCl}=0.089mol`

Now we have to calculate the number of moles of HCl for trial 3.

Volume of HCl = 15.2 mL = 0.0152 L

`6.0M=\frac{\text{Moles of HCl}}{0.0152L}`

`\text{Moles of HCl}=0.091mol`