Question

I would like help understanding these questions.

3. Calculate the mass of sodium chloride needed to generate 0.750 L of a 2.50 M solution of sodium chloride. 4. A 1.00 L stock solution of 6.00 M sulfuric acid is diluted to a new volume of 3.00 L. Calculate the molarity of the new solution. 5. Calculate the molarity of a new solution when a 750. mL stock solution of 12.0 M hydrochloric acid has 250. mL of water added to it.

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Answer 1

To find the mass of sodium chloride (Naa) Given that 0.750 L of a 2.50 M solution of sodium chloride First find the moles of sodium chloride moles = molarity x volume = 2.50mol/L 0.750L = 1.875moles To find the mass of sodium chloride (NaCl) mass = moles Xmolarmass molarmass of NaCl = 58.44g/mol mass = 1.875x58.44g/mol = 109.57 g Hence the mass of sodium chloride (Naa)= 109.57 g

To find the molarity of the new solution Given that 1.00L of 6.00M sulfuric acid is diluted to 3.00L We have to use the titration formula MV = MV M = Initial molarity = 6.00M Vi = Initial volume =1.00L My = Final molarity = ? V = Final volume = 3.00L MV = M,V 12 V2 M,M,L M. - 6.00Mx1.00L M2= 3.00L M, = 2M Hence the the molarity of the new solution = 2M (5) To find the molarity of the new solution Given that 750 mL of 12.0M hydrochloric acid has 250mL of water We have to use the titration formula MV=M,V2 M = Initial molarity = 12.0M V1 = Initial volume = 750 mL My = Final molarity = ? V = Final volume = 250mL MV = MV M2=72 M - 250mL 12.0Mx 750 mL My = 36M Hence the the molarity of the new solution = 36M