Question

You need to prepare 150 mL of 0.1 M solution of silver chloride. How much silver chloride is required?

Answer 1

amount of silver chloride required is 0.015 moles or 2.1504 g

Step-by-step explanation:

0.1M AgCL means 0.1mol/dm³ or 0.1mol/L

1L = 1000mL

if 0.1mol of AgCl is contained in 1000mL of solution

then x will be contained in 150mL of solution

cross multiply to find x

x = (0.1*150)/1000

x= 0.015 moles

moles of silver chloride present in 150 mL of solution is 0.15 moles

To convert this to grams, simply multiply this value by the molar mass of silver chloride

molar mass of silver chloride AgCl =107.86 + 35.5

=143.36 g/mol

mass of AgCl = moles *molar mass

=0.015*143.36

=2.1504g

=