Answer:
Mass = 12.27 g
Explanation:
Given data:
Mass of NaCl are required = ?
Volume of AgNO₃ = 346 mL (346 mL× 1L/ 1000 mL = 0.346 L)
Molarity of AgNO₃ = 0.607 M
Solution:
Chemical equation:
NaCl + AgNO₃ → NaNO₃ + AgCl
Number of moles of AgNO₃:
Molarity = number of moles / volume in L
0.607 M = number of moles / 0.346 L
number of moles = 0.607 M × 0.346 L
number of moles = 0.21 mol
now we will compare the moles of NaCl and AgNO₃.
AgNO₃ : NaCl
1 : 1
0.21 : 0.21
Mass of NaCl:
Mass = number of moles × molar mass
Mass = 0.21 mol × 58.44 g/mol
Mass = 12.27 g
