A0.3146-g sample of a mixture of nacl(s) and kbr(s) was dissolved in water. the resulting solution required 45.30 ml of 0.08765 m agno3(aq) to precipitate the cl–(aq) and br–(aq) as agcl(s) and agbr(s). calculate the mass percentage of nacl(s) in the mixture.

Let mass of NaCl in the mixture = x g

Mass of KBr = 0.3146 -x

Moles of NaCl = Mass / Molar mass = x / 58.5

Mole sof KBr = 0.3146-x / 119

Moles of AgNO3 added = molarity X volume (L) = 0.08765 X 45.30 / 1000 = 0.00397 moles

Out of these say moles of AgNO3 reacted with Cl- = y

So moles of AgNO3 reacted with Br- = 0.00397- y

We know that

x / 58.5  = y  (1)

0.3146-x / 119 = 0.00397 - y (2)

Putting (1) in (2)

0.3146 -x /119 = 0.00397 - x / 58.5

18.40 - 58.5x = 27.64 - 119x

x = 0.153 grams  = Mass of NaCl

MAss % of NaCl  = Mass of NaCl X 100 / total mass

                  = 0.153 X 100 / 0.3146 = 48.63%

i believe it is c.) 1575 j.

idk what the question is about