Answer:
Molarity of the solution = 3.000 M
Volume of the solution = 250.0 mL = 0.25 L
moles in 250.0 mL = molarity x volume of the solution
= 3.000 M x 0.25 L
= 0.75 mol
Hence, 0.75 mol of NaCl is needed to prepare 250.0 mL of 3.000 M NaCl solution.
Moles (mol) = mass (g) / molar mass (g/mol)
Moles of NaCl in 250.0 mL = 0.75 mol
Molar mass of NaCl = 58.44 g/mol
Mass of NaCl in 250.0 mL = Moles x Molar mass
= 0.75 mol x 58.44 g/mol
= 43.83 g
Hence, 43.83 g of NaCl is needed to prepare 250.0 mL of 3.000 M solution.
Explanation:
Answer:
17.04%
Explanation:
Actual Value = 173.1
Measured Value = 143.6
Percent error is obtained using the equation;
Percent error = (Measured - Actual) / Actual ]* 100
Percent error = [ (143.6 - 173.1) / 173.1 ] * 100
The absolute value of (Measured - Actual) is taken,
Percent Error = [29.5 / 173.1 ] * 100
Percent Error = 0.1704 * 100 = 17.04%
If there is 4 of CH2CI2, then there is 8 of CI. There is already 2 in each one and there is 4 sets so 2x4=8
