Question

The concentration of Cl- ions in a mixture obtainedby mixing 100mL of 0.1 M NaCl with 40 mL of 0.1M AgNO3, is:
please give answer with proper explanation​

Answer 1

Molarity of NaCl = 0.1 M

Volume = 100 ml = 0.1 L

We know that ,

Molarity = moles of NaCl / Volume

moles of NaCl = 0.1 × 0.1 = 0.01 moles

moles of NaCl = 0.01 M

no of moles of Cl - ions = 0.1× 1 = 0.1

(as 1 mole NaCl gives 1 mole of Cl - ion)

Conc of Cl - ions in the soln depends only upon NaCl

So the concentration of Cl - ions in the mixture will be

= moles of Cl - / Vol of soln

total volume =100+40=140 ml =0.14 L

= 0.1 / 0.14

=0.7 M