titration question?

Question

stomach acid is HCl. 5. A sample of gastric juice having a volume of 5.00 mL required 11.00 mL of 0.0100 M KOH solution for neutralization in a titration. what was the molar concentration of HCl in this fluid? If we assume a density of 1.00 g mL^-1 for the fluid, what was the percentage by weight of HCl?

I get how to find the molar concentration of HCl but i didn't get the 2nd part of the question about the percentage by weight. i am confused by the 1.00 g mL^-1. does anyone know how to solve it? the answer is supposed to be 0.0803%

Answer 1

Yup i got the answer,

the answer for the first part should be 0.022M

Since the mole ratio of HCl : KOH is 1:1, there should be 0.00011moles of HCl present as 0.00011moles of KOH is present.

1.00g mL^-1 means that 1gram of HCl should be present in 1mL of a HCl solution, but this isnt the case.

Since 0.00011 moles of HCl is present, we can find the mass of HCl present in 5.00mL of the gastric juice:

0.00011 * ( 35.5+1)= 0.004015g

5mL of the gastric juice is present, so the actual mass of HCl present should be 5grams. However only 0.004015grams of HCl are present, so the percentage by weight of HCl can be found simply by

(0.004015/ 5) * 100 = 0.0803%

taadaa~!

Answer 2

You are given how much matter occupies space by the density. So if 1 mL of space contains 1.00g of matter then 5mL contain 5g of matter (5mL * 1g/mL) Since you know how to find how many moles of HCl in the 5mL of space, convert that answer into grams and find the percentage of weight. (part/whole * 100)
0.00011mol HCl *(36.458g HCl/mol HCl) = 0.00401 g HCl
(0.00401g HCl/5g Solution) * 100 = 0.0802%

Answer 3

A titration is used while a answer of common concentration could be added to and react with yet another fabric it is of unknown concentration. the quantity of the addion of the answer has to have the skill to be measured. The common concentration fabric could react in a typical proportion to the unknown fabric so as that the quantity of the unknown could be desperate. There could be some thank you to be responsive to while the unknown has been totally fed on so as that greater of the familiar answer isn't added. permit's say the reaction of the familiar, ok to the unknown, U is two:a million (2 moles ok reacts with a million mole U). permit's assume that the familiar concentration is 0.a million M and that we devour 35 ml of the answer in the titration. this suggests that we fed on 35 ml X 0.a million mmole/ml = 3.5 mmole of ok. because of the fact of the two:a million ratio, this suggests there's a million.seventy 5 mmoles of U in the pattern. Titration is used while there's a sparkling, common molar ratio of common to unknown, while there is the thank you to verify end factor, and while the reaction is speedy.