<span> mass of glucose = 0.055 *165 = 9.075 g
vol of methyl alc = 0.185 * 1.87 = 0.346 L = 346 ml
% NaCl ( m/v ) = mass NaCl * 100/ vol of soln
or Vol of Soln = mass NaCl / % NaCl (m/v)
= 32.1 * 100 / 6 = 535 ml the total vol of soln</span>
Answer :
The concentration of
before any titrant added to our starting material is 0.200 M.
The pH based on this
ion concentration is 0.698
Explanation :
First we have to calculate the concentration of
before any titrant is added to our starting material.
As we are given:
Concentration of HBr = 0.200 M
As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion
and bromide ion
.
As, 1 M of HBr dissociates to give 1 M of 
So, 0.200 M of HBr dissociates to give 0.200 M of 
Thus, the concentration of
before any titrant added to our starting material is 0.200 M.
Now we have to calculate the pH based on this
ion concentration.
pH : It is defined as the negative logarithm of hydrogen ion concentration.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)


Thus, the pH based on this
ion concentration is 0.698
First one??
I believe this is the correct answer
Answer:
molecular formulas of the hydrocarbons