Question

17. What volume of 0.345 M NaCl solution is needed to obtain 10.0 g of NaCl?

18 How many milliliters of a 0.0015 M cocaine hydrochloride (C₁₇H₂₂ClNO₄) solution is needed to obtain 0.010 g of the solute?

19.. A solution having an initial concentration of 0.445 M and the initial volume of 45.0 mL is diluted to 100.0 mL. What is its final concentration?

20. A 50.0 mL sample of salt water that is 3.0% m/v is diluted to 950 mL. What is its final mass/volume percent?

Answer 1

Given Concentration of solution =.= 0.345M Concentration - No of moles V = Volume of solution in Litree 0.345M = 2 0.345M = M = Molar mass of Nach = m = 10-08 58.5g/mole 0.1709 mole 0.345M = 0.17ogmole [V = 0.495. v Answer V = Total volume solution in Litra. 18 of Concontration Molar mass of Concentration = 2 Coccine hydrochloride =( = ONISM Cocaine hydrochloride = M = 339.123 g/mol = No of molu Volume of solution = orolog 2.948 10 mole 339.123g/mol n = Given mass Molar mass 뀨 = l = V 19-60 X 10 L - Th V = 19.60mL 2.94x10s mole 0.0OISM 1000mL Ang

Initial concentration of solution Initial volume = v = 45.0mL -c, = 0.445M final volume of Suppose final solution = v concentration = loomL of solution beton av =9Vz 10.445 MX 45.0mL = 5x loomL s = 0.200M / Answer