What is the molarity of the solution formed when 7.88 grams of nacl is mixed with enough water to make 350. ml of solution?

The molarity of the solution is 0.386 M

Explanation:

Step 1: Data given

Mass of NaCl = 7.88 grams

Volume of the solution = 350 mL

Molar mass of NaCl = 58.5 g/mol

Step 2: Calculate number of moles

Number of moles NaCl = mass of NaCl / molar mass of NaCl

Number of moles NaCl = 7.88 grams / 58.5 g/mol

Number of moles = 0.135 moles

Step 3: Calculate molarity of the solution

Molarity = Number of moles NaCl / volume

Molarity = 0.135 moles / 0.350 L

Molarity = 0.386 M

The molarity of the solution is 0.386M

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