Answer:
8.77g
Explanation:
Step 1:
Data obtained from the question.
This includes the following:
Concentration of A (C1) =.?
Volume of A (V1) = 16 mL
Volume of B (V2) = 300 mL
Concentration of B (C2) = 0.50 M
Molar Mass of NaCl = 58.443 g/mol
Mass of NaCl =.?
Step 2:
Determination of the concentration of A.
Applying the dilution formula:
C1V1 = C2V2
The concentration of A i.e C1 can be obtained as follow:
C1V1 = C2V2
C1 x 16 = 0.5 x 300
Divide both side by 16
C1 = (0.5 x 300) / 16
C1 = 9.375 M
Therefore, the concentration of A is 9.375 M
Step 3:
Determination of the number of mole of NaCl in 9.375 M NaCl solution. This is illustrated below:
Molarity = 9.375 M
Volume = 16 mL = 16/1000 = 0.016 L
Mole of NaCl =?
Molarity = mole /Volume
Mole =Molarity x Volume
Mole of NaCl = 9.375 x 0.016
Mole of NaCl = 0.15 mole
Step 4:
Determination of the mass of NaCl. This is illustrated below:
Mole of NaCl = 0.15 mole
Molar Mass of NaCl = 58.443 g/mol
Mass of NaCl =?
Mass = number of mole x molar Mass
Mass of NaCl = 0.15 x 58.443
Mass of NaCl = 8.77g
Therefore, 8.77g of NaCl is needed to make 1 L of the original solution A.
