Pure acetic acid has a density of 1.049g / (ml) . calculate the molarity of a solution of acetic acid made by dissolving 20.00 ml in enough water to make 250.0 ml of solution

Molarity = 1.4 M

Explanation:

Given data:

Density of acetic acid = 1.049 g/mL

Volume of acetic acid = 20.0 mL

Volume of solution = 250.0 mL (250/1000 = 0.25 L)

Molarity = ?

Solution:

Mass of acetic acid = density × volume

Mass of acetic acid = 1.049 g/mL × 20.0 mL

Mass of acetic acid = 20.98 g

Number of moles of acetic acid = mass/ molar mass

Number of moles of acetic acid  =  20.98 g/ 60.052 g/mol

Number of moles of acetic acid  = 0.35 mol

Molarity = number of moles / volume in L

Molarity = 0.35 mol / 0.25 L

Molarity = 1.4 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

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