Question

3.355 M K2SO4 solution can be prepared by A) dilution of 500.0 mL of 1.00 MK2SO4 to 1.00 L B) dissolving 46.7 g of K2SO4 in water and diluting to 500.0 ml, then diluting 25.0 ml of this solution to a total volume of 500.0 mL C) dissolving 46.7 g of K2SO4 in water and diluting to a total volume of 250.0 mL D) dilution of 1.00 mL of 0.355 M K2SO4 to 1.00 L E) diluting 46.7 mL of 1.90 MK2SO4 solution to 250,0 mL which solution has the same number of moles of HCl as 25.0 ml of 0.200 M solution of HC A) 25.0 ml of 0.185 M solution of HCI B) 45.0 mL of 0.155 M solution of HCI C) 100.0 ml of 0.500 M solution of HCI D) 12.5 mL of 0.400 M solution of HCI E) 500 mL of 0.150 M solution of HCI 3 What are the respective concentrations (M) of Fe3+ and l-afforded by dissolving 0.200 mol Fels in water and diluting to 725 mL? A) 0.828 and 0.276 B) 0.276 and 0.828 C) 0.276 and 0.276 D) 0.145 and 0.0483 E) 0.145 and 0.435 Which of the following 0.300 M solutions would contain the highest concentration of potassium ions? A) potassium hydrogen carbonate B) potassium hypochlorite C) potassium phosphate D) potassium iodide E) potassium bromide A 650 mL sodium bromide solution has a bromide ion concentration of 0.245 M. What is the mass (s) of sodium bromide in solution? A) 0.159 B) 164 0.00155 D) 103 E) 16400 5) What volume (L) of 0.250 M HNO is required to neutralize a solution prepared by dissolving 17.5 g of NaOH in 350 ml of water? A) 1.75 B) 1.25 1.75 10-3 D) 0.11 E) 0.070 6) An aliquot (287 mL) of a KOH solution required 31.3 ml of 0.118 M HCl for neutralization. What mass (g) of KOH was in the original sample? A) 0.173 B) 0.129 0.215 D) 0.190 E) 0.207

Answer 1

- Required Tils Given Concentration of ki so = c = 0.355M y connect La volume of Kosoa initially = V = 500m = 0.5L IL = looon Concentration of k₂so -c = 1.oom tinal volumelval = 1 Dol cv = Cava Cuc = final concentration of Ksa 1.OOM X 0.5L = X IL a = 0.5M Answer - This option is in connect (( Amount of kiso = 46.79 Molar mass of K2504 = 174.259 g/mol. No of moles of kson - (46.79 174.259 g/mole) 0.268 mod - - 250A = 0.268 mole - Volume = v= 5oomL = 0.5l Concentration $ 0.536M 0.5L No. of males of Kesen in 25ml = 0.53614 X 0.025L - .0134 mole l volume = v= 500ml v = o-SL Concentration of kis04 = 0.0134mole 0.0268M 0.5L Amount of kaso = 46.7g Molar mass of ks = 174.259 g/mol No of mole of Kiso = 46.79 174.259 g/mol Total volume = 250mL = 0.250L Concentration of K₂SO4 = 0.268 mole z 1-0729 H -2 4- 0.2501 e = 0.268 Mole 0.250L

D 0 = 0.Doll _ Initial concentration of ks 04 = V: =l.com Initial volume of kaso = V = 1.00mL V Concentration of Kas4 = G = 0.35SM. final volume = v = 100L cv = CV2 6 = final concentration of kson - 0,355M X o.oolL = 5 x 1.ool = 0.0003 SSM I > Incorrect E Initial volume = V = 46.4mL = 0.0467L Initial concentration = c = 1.90m final volume = V₂ = 250.0mL - V₂ = 0.asol cv = va s = final concentration of kason 1.90M X 0.0467L = 6 x 0.250L Ig = 0.355M Answa Connect