What is the final concentration if water is added to 0.25 L of a 8M NaOH solution to make 4.0 L of a diluted NaOH solution? .25 M NaOH solution
1 M NaOH solution
.5 M NaOH solution

0.5 M NaOH solution

Explanation:

Initially, we have 0.25L or 250 mL of 8M NaOH solution

Which means that 1L of this solution would have 8 moles of NaOH

Moles present in 250 mL of solution:

Molarity * Volume( in L)

8 * 0.25 = 2 moles

There is 2 moles of NaOH in 0.25 L of the provided NaOH solution

More water is going to be added to this solution but the number of moles of NaOH will remain the same

now that the solution has a volume of 4L after adding the additional water, we have a 4L solution which contains 2 moles of NaOH

Molarity of the new solution:

Molarity = Number of moles in solution / Volume of solution

Molarity = 2 / 4

Molarity = 0.5 M

where's the diagram?

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