Asample of seawater contains 78.0 grams of nacl in 2,025 ml of solution. if the molar mass of nacl is 58.443 g/mol, what is the molarity of this sample?

Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically,        Molarity =

Also, it is known that no. of moles equal mass divided by molar mass. Since, it is given that 2,025 mL equals 2.025 L, mass is 78.0 g, and molar mass is 58.443 g/mol.

Therefore,     No. of moles =

                                            =

                                            = 1.33 mol

Hence, calculate the molarity as follows.

                 Molarity =

                                =

                                = 0.656 mol/L

Thus, we can conclude that molarity of the given sample is 0.656 mol/L.

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

the right answer is c i just did the question.