Title: plz i need some help in solving some problems in chemistry 101 Post by: luluwa_999 on Nov 4, 2012 I want some help in these Question :
1- Calcium nitrate tetrahydrate dissolves in cold water to the extent of 266 g per 100. cm3. What is the concentration of nitrate ions in this solutio n? 2- 34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H2SO4 solution. What is the concentration of the original sulfuric acid solution? 3- 35.0 mL of 0.255 M nitric acid is added to 45.0 mL of 0.328 M Mg(NO3)2. What is the concentration of nitrate ion in the final solution? 4- What mass of K2CO3 is needed to prepare 200. mL of a solution having a potassium ion concentration of 0.150 M? 5- What is the molar concentration of chloride ions in a solution prepared by mixing 100. mL of 2.0 M KCl with 50. mL of a 1.5 M CaCl2 solution? Thank you Title: Re: plz i need some help in solving some problems in chemistry 101 Post by: jerica.baino on Nov 4, 2012 1- Calcium nitrate tetrahydrate dissolves in cold water to the extent of 266 g per 100. cm3. What is the concentration of nitrate ions in this solutio n? moles Ca(NO3)2 * 4 H2O = 266 g / 236.15 g/mol=1.13 moles NO3- = 2 x 1.13 = 2.26 concentration NO3- = 2.26 / 0.100 L = 22.6 M Title: Re: plz i need some help in solving some problems in chemistry 101 Post by: jerica.baino on Nov 4, 2012 2- 34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H2SO4 solution. What is the concentration of the original sulfuric acid solution? 2NaOH + H2SO4 ----> Na2SO4 + 2H2O number of mole of NaOH=(0.1510 X 34.62)/1000=5.23X10^ -3mol number of mole of H2SO4=(5.23X10^ -3)/2=2.61X10 ^-3mol original volume of H2SO4=50ml=0.05L original concentration of H2SO4=2.61X10 ^-3mol/0.05L=0.0522mol/L Title: Re: plz i need some help in solving some problems in chemistry 101 Post by: bio_man on Nov 4, 2012 3- 35.0 mL of 0.255 M nitric acid is added to 45.0 mL of 0.328 M Mg(NO3)2. What is the concentration of nitrate ion in the final solution? Multiply .255 by .0350 L to get .00893 moles HNO3 (sig figs) Multiply .328 by .0450 L to get .0148 moles Mg(NO3)2 (sig figs). However, there are .0296 moles of nitrate ions. .00893 mol + .0296 mol = .0385 mol (sig figs) .0350 L + .0450 L = .0800 L .0385 mol / .0800 L = .481 M |