Achemist needs to know the concentration of some naoh that is in the laboratory. to find this out, she titrates a 50.0 ml sample of the solution with 3.5 m hcl. if it takes 34.3 ml of the hcl to reach the titration endpoint, what is the concentration of the naoh solution?

2.4 M

Explanation:

According to the neutralization law,

where,

= molarity of solution = 3.5 M

= volume of solution = 34.3 ml

= molarity of solution = ?

= volume of solution = 50 ml

= valency of = 1

= valency of = 1

Therefore, the concentration of NaOH solution is 2.4 M.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g