Problem #1 (a) What is the concentration (expressed as mol/L = M) of NaClwhen 32

Question

Problem #1 (a) What is the concentration (expressed as mol/L = M) of NaClwhen 32.0g of NaCl is dissolved in water and diluted to0.500L? (b) How many grams of methanol (CH3OH, formulaweight = 32.04) are contained in 0.100L of 1.71M aqueousmethanol? (c) Any dilute aqueous solution has a density near 1.00 g/mL.Suppose the solution contains 1 ppm of solution ; express theconcentration of solution in g/L, mg/L, g/mL, and mg/mL. Problem #1 (a) What is the concentration (expressed as mol/L = M) of NaClwhen 32.0g of NaCl is dissolved in water and diluted to0.500L? (b) How many grams of methanol (CH3OH, formulaweight = 32.04) are contained in 0.100L of 1.71M aqueousmethanol? (c) Any dilute aqueous solution has a density near 1.00 g/mL.Suppose the solution contains 1 ppm of solution ; express theconcentration of solution in g/L, mg/L, g/mL, and mg/mL.

Explanation / Answer

(a) Given mass , m = 32 g V = Volume = 0.5 L Molar mass of NaCl , M' = 23 + 35.5 = 58.5 g V = Volume = 0.5 L Molar mass of NaCl , M' = 23 + 35.5 = 58.5 g Molarity , M = no. of moles / Volume in L                     = ( mass / Molar mass ) / Volume in L                     = ( 32 / 58.5 ) / 0.5                     = 1.094 mol / L (b)Given mass , m = ? V = Volume = 0.1 L Molar mass of CH3OH =32.04 g Molarity , M = 1.71 M Molarity , M = no. of moles / Volume in L                     = ( mass / Molar mass ) / Volume in L 1.71 = ( m / 32.04 ) / 0.1 m = 5.4788 g (c)1ppm = 1mg /L V = Volume = 0.1 L Molar mass of CH3OH =32.04 g Molarity , M = 1.71 M                     = ( mass / Molar mass ) / Volume in L 1.71 = ( m / 32.04 ) / 0.1 m = 5.4788 g (c)1ppm = 1mg /L

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