0.850g Silver nitrate are dissolved in 250cm3 water. calculate the
molarity of the solution.
The salt from a 25g packet of crisps is extracted with water and made up to 50cm3.
20cm3 of this required 11.2cm3 of silver nitrate to complete the reaction.
Calculate the % by wt of salt in the crisps

Hello Paulinej

The molar mass os Silver Nitrate (AgNO3) is: 107.868+14.007+47.997 = 169.872 g/mol

The number of mol of AgNO3 in 0.850g is: 0.850g / 169.872 g/mol = 0.005 mol

The molarity of the solution is: 0.005 mol / 0.250 L = 0.02 M (or mol / L).

The second part of your problem is this:

NaCl + AgNO3 -> NaNO3 + AgCl

This means that one mol of NaCl reacts with one mol of AgNO3 to form
the precipitate.

We can calculate the number of mol of AgNO3 used as this:

(11.2/1000)L x 0.02 M = 0.00024 mol

The number of mol of NaCl used was therefore also 0.00024 mol.  This
was contained within 20 cm3 of water.  The molarity of this solution
was therefore:

0.00024 mol / (20/1000) = 0.0112 mol/L

We originally had 50 cm3 of water in the solution therefore the total
number of moles in that solution is 0.0112 mol/L * (50/1000) = 0.00056
mol.

The molar mass of NaCl is 22.98977 + 35.453 = 58.443 g/mol

Therefore the amount of salt in the crisp solution is 0.00056 mol x
58.443 g/mol = 0.03275g

The % by weight of salt in the crisps is (0.03275g / 25g) x 100% = 0.13%

You may want to double-check all the numbers here to fully understand
what is going on, to alter the number of significant figures required
and to verify my solution.