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Welcome to the World of MOLARITY

Welcome to the World of MOLARITY. We may now explore the different sample problems given. BACK. NEXT. MOLARITY. Molarity expresses the concentration of a solution in number of moles of solute per liter of solution. Mathematically; M=n/V

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Welcome to the World of MOLARITY

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  1. Welcome to the World of MOLARITY • We may now explore the different sample problems given. BACK NEXT

  2. MOLARITY Molarityexpresses the concentration of a solution in number of moles of solute per liter of solution. Mathematically; M=n/V where M=molarity (in moles solute/liter solution) n=number of moles of solute V=volume of solution (in liters)

  3. SAMPLE PROBLEM # 1 A solution of sodium hydroxide was prepared by dissolving 10.0 grams of sodium hydroxide (NaOH) in distilled water making the volume up to 500 cm3. What is the molarity of the NaOH solution? Atomic masses: Na=23; O=16, H=1 GIVEN: mass of solute (NaOH)= 10.0 grams molar mass of NaOH=1(23)+1(16)+1(1) =40g/mole n (moles of NaOH)= 10.0g(1mol/40g)=0.25moles V of NaOH solution= 500cm3(1L/1000cm3)=0.50L BACK NEXT

  4. Solution to SP # 1 GIVEN: mass of solute (NaOH)= 10.0 grams molar mass of NaOH=1(23)+1(16)+1(1) =40g/mole n (moles of NaOH)= 10.0g(1mol/40g)=0.25moles V of NaOH solution= 500 cm3(1L/1000cm3)=0.50L SOLUTION: M(NaOH solution)=n(moles NaOH)/V(volume solution in L) =0.25moles /0.50L = 0.5 M or 0.5 moles/liter A 0.5 M solution, means there are 0.5 moles of NaOH in 1 liter of NaOH solution BACK NEXT

  5. SAMPLE PROBLEM # 2 Calculate the number of moles copper sulfate in a 250 milliliter (mL) of 0.1 molar copper sulfate (CuSO4) solution. GIVEN: M =0.1moles/L V of CuSO4 solution= 250mL(1L/1000mL)=0.25L n (moles of CuSO4)= ? BACK NEXT

  6. Solution to SP # 2 GIVEN: M =0.1moles/L V of CuSO4 solution= 250mL(1L/1000mL)=0.25L n (moles of CuSO4)= ? SOLUTION: n(moles CuSO4)=(V of CuSO4 solution in L)x (M of CuSO4 solution) =(0.25 L) (0.1 moles/ L) = 2.5 molesCuSO4 There are 2.5 moles of CuSO4 in a 250 mL of 0.1 M CuSO4 solution BACK NEXT

  7. SAMPLE PROBLEM # 3 What is the volume of a 0.50 molar potassium bromide (KBr) solution containing 1.5 moles of potassium bromide (KBr)? GIVEN: M =0.5 moles/L means there are 0.5 moles KBr in 1L KBr solution n =1.5 moles KBr V of KBrsolution=? BACK NEXT

  8. Solution to SP # 3 GIVEN: M =0.5 moles/L means there are 0.5 moles KBr in 1L KBr solution n =1.5 moles KBr V of KBrsolution=? SOLUTION: V of KBrsolution in L) =n(moles KBr)/( M of KBr solution) (1.5 moles) (0.5 moles/ L) = = 3.0 L KBr BACK NEXT

  9. SAMPLE PROBLEM # 4 3. A 250 milliliter bottle of vinegar reads 5% acidity (5 g acetic acid in 100 gram vinegar).Calculate the concentration of acetic acid in moles per liter. (1mole CH3COOH=60 grams) (100mL) (60g acetic acid) GIVEN: V of solution =250 mL n acetic acid=250mL (5g acetic acid)(1mole acetic acid) =0.208 moles acetic acid M=0.208 moles 0.25 L M=0.83 moles/liter BACK NEXT

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  11. SOLUTIONS AT WORK Sheet #1 A solution of glucose (C6H12O6) contains 300 grams in 1.0 liter of solution. What is the concentration of the glucose solution in moles per liter? Atomic masses: C=12, H=1, O=16 BACK NEXT

  12. SOLUTIONS AT WORK Sheet #1 2. A 3.0 grams of magnesium hydroxide, Mg(OH)2 is dissolved in water to produce a 0.2 molar magnesium hydroxide solution. What is the volume of the solution prepared? Atomic masses: Mg=24, H=1, O=16 BACK NEXT

  13. Assignment: Answer the following briefly: There are times when glucose solution must be introduced into the bodies of patient. Why is it dangerous for this solution to be highly concentrated? BACK HOME NEXT

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