Question

erksheet 7-Rey 10 A 10.0 mL sample of sodium chioride solution that has a mass of 16.78 g is placed in an Solutions evaporating dish and evaporated to dryness. The residue has a mass of 6.55 g. Calculate the following concentrations for the NaCl solution mass/mass (m/m) percent mass/volume (m/v) percent malarity A 4.00 % (g/mL) KI solution has a volume of 65.0 mL. How many grams of KI are in the sample? How many grams of a 35% ( m/m ) NaCI solution contain 250.0 g of NaC? What is the molarity of a solution that contains 50.0 g of NaOH (molar mass, 40.0 g/mol) dissolved in 500.0 mL of solution? How many milliliters of a 1.75 M MgCl2 solution contain 32.5 g of MgCl2? If 10.0 mL of solution A is diluted to 50.0 mL, what is the concentration of the final solution (solution B)? Consider a 1.00 M CaCl2 solution (called solution A): If 25 mL of the initial solution (solution A) is diluted to 500. mL (solution C) and then 37 mL of solution C is diluted to 250 mL resulting in solution D, what is the concentration of solution D?

Answer 1

0 Here volume of solution = lomL mass of solution = 16.78 g mass of Residue (Desy Mace after evaporation - 6.55g Here 6.55 g af Mace is present in 16.70g af solution so concentration mass/mars a mass of Mace Dry X100 ( Mas of solution 6.55 X 100 = 16.78 (ii concentration in mlva = 39.03% 9/g mass of Dry Mace x100 volume of solution (me 6.5 x 100 = [65.5% / mL 10 me

Now Molarity di Here Dry Macea 6.55g molar mabof Mace = 35.5+23 F = 58.5 g/mol moles of Mace in 6.13 g = 6.3 = S8.5 =0.112 mole volume of solution a 10mL = 0.01L 1000mL 2 12 so lomba Ooll Mow Molarity a moleof Mall Volume of suutios in L. con el - 0.12 - 11.2 Molar) 0.01 Here 4.0% g/mL concentration given This means that in 100 mhof solution there is Yg of ks present & we need to find amount of FI in 65 g me solution so : 100m 65 we posses KI = ug will posses a 4465 196 al