Question

0.12g of rock salt was dissolved in water and titrated with 0.1moldm^-3 silver nitrate until the first permanent brown precipitate of silver chromate is seen. 19.70 cm^3 was required to titrate all the chloride ion. how many moles of chloride ion were titrated? what mass of sodium chloride was titrated? what was the % purity of the rock salt in terms of sodium chloride?

Answer 1

moles Cl⁻ : 0.00197

mass NaCl : 0.115

% purity : 95.83%

Further explanation

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

`\large \boxed {\bold {M ~ = ~ \frac {n} {V}}}`

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

So to find the number of moles can be expressed as

n = V x M

• mol Cl⁻

Reaction

AgNO₃ + NaCl ⇒ AgCl + NaNO₃

Molarity(concentration) of AgNO₃ = 0.1 mol/dm³(L) = 0.1 M

Volume=V of AgNO₃ = 19.7 ml(cm³)

so mol of AgNO₃ :

`\tt mol=M* V\\\\mol=0.1~mol/L* 0.0197~L\\\\mol=0.00197`.

From the equation, mol ratio AgNO₃ : NaCl = 1 ; 1, so mol NaCl= mol AgNO₃= 0.00197

NaCl⇒Na⁺+Cl⁻

mol Cl⁻ : mol NaCl = 1 : 1, mol Cl⁻ = 0.00197

• mass NaCl (MW=58.5 g/mol) :

`\tt mass=mol* MW\\\\mass= 0.00197* 58.5=0.115~g`

• % purity :

`\tt \%purity=(mass~NaCl)/(mass~rock)* 100\%\\\\\%purity=(0.115)/(0.12)* 100\%=95.83\%`