1] a) The concentration of C 32 H 66 inwinter rainwater is 1.50 ppb. Assume that
ID: 76911 • Letter: 1
Question
1]a) The concentration of C32H66 inwinter rainwater is 1.50 ppb. Assume that the density of water is1.00 g/mL; find the concentration of C32H66expressed as nM. Hint: If % = parts per 100, then what would ppb orparts per billion equal? (molecular weight C32H66 = 450.88)
b)The concentration of sugar (glucose,C6H12O6) in human blood rangesfrom about 80 mg/100 mL before meals up to 120 mg/100 mL aftereating. A certain blood sample had a glucose concentration of 97mg/100 mL. Express the concentration in mM. (molecular weightglucose = 180.16)
c) It is recommended that drinking water contain 1.6ppm fluoride (F-) for prevention of tooth decay. Acertain reservoir contained an estimated 1.67 x 1012 mLwater. How many kg of NaF should be added to give 1.6 ppmF-? Hint: If % = parts per 100, then what would ppm orparts per million equal? (formula weight NaF = 41.98817)
d)How many grams of NaF are required to prepare500 mL of 0.19 M F-? (formula weight 41.98817)
e)How many grams of Potassium Hydrogen Phthalate(KHP), are required to make 500 mL of 0.48 M KHP? (formula weight= 204.23)
Explanation / Answer
a) The concentration of C32H66 in winterrainwater is 1.50 ppb. Assume that the density of water is 1.00g/mL; find the concentration of C32H66expressed as nM. Hint: If % = parts per 100, then what would ppb orparts per billion equal? (molecular weight C32H66 = 450.88)Concentration of C32H66 in winter rainwateris 1.50 ppb 1 ppb = 0.001 ppm Concentration = 1.5 ppb = 1.5 * 0.001 ppm = 0.0015ppm 1 ppm = 1 mg / L 0.0015 ppm = 0.0015 mg / L Molarity = ( wt / Mol.wt. ) / Volume in L = ( 0.0015 / 450.88 ) / 1 L = 3.3268 * 10 -6 M = 3.3268 * 10 -6 * 10 -3 * 10 3M = 3.3268 * 10 3 * 10 -9 M =3.3268 * 10 3nM Since 1nM = 10 -9 M
b)The concentration of sugar (glucose,C6H12O6) in human blood rangesfrom about 80 mg/100 mL before meals up to 120 mg/100 mL aftereating. A certain blood sample had a glucose concentration of 97mg/100 mL. Express the concentration in mM. (molecular weightglucose = 180.16) Molarity = ( wt / Mol.wt. ) / Volume in L = ( 0.097 / 180.16 ) / 0.1 L = 5.3841 * 10 -3 M =5.3841 mM
d)How many grams of NaF are required to prepare500 mL of 0.19 M F-? (formula weight 41.98817) Molarity = ( wt / Mol.wt. ) / Volume in L 0.19 M= ( w / 41.98817g ) / 0.5 L w = 3.9888 g
e)How many grams of Potassium Hydrogen Phthalate(KHP), are required to make 500 mL of 0.48 M KHP? (formula weight= 204.23)
Molarity = ( wt / Mol.wt. ) / Volume in L 0.48 M= ( w / 204.23 g) / 0.5 L w = 49.0152 g
Molarity = ( wt / Mol.wt. ) / Volume in L = ( 0.097 / 180.16 ) / 0.1 L = 5.3841 * 10 -3 M =5.3841 mM
d)How many grams of NaF are required to prepare500 mL of 0.19 M F-? (formula weight 41.98817) Molarity = ( wt / Mol.wt. ) / Volume in L 0.19 M= ( w / 41.98817g ) / 0.5 L w = 3.9888 g
e)How many grams of Potassium Hydrogen Phthalate(KHP), are required to make 500 mL of 0.48 M KHP? (formula weight= 204.23)
Molarity = ( wt / Mol.wt. ) / Volume in L 0.48 M= ( w / 204.23 g) / 0.5 L w = 49.0152 g
Molarity = ( wt / Mol.wt. ) / Volume in L 0.19 M= ( w / 41.98817g ) / 0.5 L w = 3.9888 g
e)How many grams of Potassium Hydrogen Phthalate(KHP), are required to make 500 mL of 0.48 M KHP? (formula weight= 204.23)
Molarity = ( wt / Mol.wt. ) / Volume in L 0.48 M= ( w / 204.23 g) / 0.5 L w = 49.0152 g
Molarity = ( wt / Mol.wt. ) / Volume in L 0.48 M= ( w / 204.23 g) / 0.5 L w = 49.0152 g
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