If 28 mL of NaCl reacts completely with 46 mL of 0.44 M AgNO3, what is the molarity of the NaCl solution? The reaction is: AgNO3 + NaCl -> AgCl + NaNO3
Since we have a balanced chemical equation, we can use the stoichiometry of the equation to solve for the answer. The first thing to look at is what we are trying to find, which is molarity of NaCl. To find this, we need moles of NaCl and L of solution, which we already know as it was given in the problem.
If we can find moles of NaCl, we can find the molarity. Since we know the mole ratio for AgNO3 to NaCl is 1:1, we know that for every one mole of AgNO3 that reacts, it will react with 1 mole of NaCl.
Given the volume and molarity of AgNO3, we can use this information to find moles, then use the coefficients for the mole ratio to find moles of NaCl
46 mL (1 L / 1000 mL) (0.44 mol AgNO3 / L soln) (1 mol NaCl / 1 mol AgNO3) = 0.020 mol NaCl
Now, we can use that value to find the molarity after converting 28 mL to 0.028 L by dividing by 1000.
0.020 mol / 0.028 L = 0.72 M NaCl
the amount of moles = volume(ml)*molarity*10⁻³
get the molarity of NaCl as x
Amount of NaCl moles = 28x*10⁻³ ⇒ ❶
Amout of AgNO₃moles = 46*0.44*10⁻³ ⇒ ❷
1AgNO₃ + 1NaCl → 1AgCl + 1NaNO₃
❶ = ❷
28x*10⁻³ = 46*0.44*10⁻³
28x = 46*0.44
x = 20.24/28
x ~ 0.722M
Hence the molarity of Nacl equals to 0.722M