# You used 1.01 g of solid and it required 15.00 ml of 1.00 M HCL to reach the endpoint. determine the no of moles of hcl in your titration?

Mar 12, 2018

You must have seen the expression..$\text{concentration"="moles of solute"/"volume of solution}$

#### Explanation:

And given the quotient....we use....

$\text{moles of solute"="concentration"xx"volume of solution}$

And so ................

${n}_{\text{HCl}} = 15.00 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 1.00 \cdot m o l \cdot {L}^{-} 1 = 0.0150 \cdot m o l$

....and thus the $1.01 \cdot g$ mass represents this molar equivalence....