Answer to Question #199253 in General Chemistry for Andre

Question

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

Answer 1

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