# Answer to Question #199253 in General Chemistry for Andre

Question #199253

Calculate the molality of a 19% HF aqueous solution. What is the ppm of HF in this solution?

Determine the mass percentage and ppm of a 0.333m NaCl solution.

Calculate the molarity of a 0.85m lithium chloride aqueous solution having a density of 1.44 g/mL

1
2021-05-27T07:33:59-0400

Calculate Molality of HF

1) Let us assume 100.0 gram of solution. Therefore:

19gm is HF

81gm is H2O

2) Molality is:

moles HF = 19gm/20.0059 g/mol = 0.95 mol

kg of water = 0.0810 kg

"molality = \\dfrac{moles\\ solute}{kilograms\\ solvent}"

mololity = 0.95 mol/0.0810 kg = 1.172 molal

ppm="\\dfrac{19}{1000}\\times 1000000"

= 19000ppm

Calculate Mass Percentage of NaCl

1) moles of NaCl Solution given = 0.333 mol

molecular weight of NaCl = 0.333 "\\times" 58.44

= 19.46 g

2) so mass percentage of NaCl is "= \\dfrac{molecular\\ weight}{Total\\ solution}"

so "= \\dfrac{19.46}{1000}\\times 100"

mass percentage= 1.946% NaCl.

ppm = "\\dfrac{19.46}{1000}\\times 1000000"

= 19460ppm

Calculate Molarity of lithium chloride

1) Molarity in terms of molality

"molarity=\\dfrac{molality\\times(density\\times1000 - molar\\ mass)}{1000}"

so, "molarity=\\dfrac{0.85\\times(1.44\\times1000 - \\ 42.394)}{1000}"

solve above equation

molarity = 1.187 M

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