Answer to Question #308894 in General Chemistry for Kev

Question

1. Consider an example in which 18.0 grams of C6H12O6 is dissolved in 1 kg of water. What is the molality of the resulting solution?

2. What is the mole fraction of sulfur dioxide (SO2) in an industrial exhaust gas containing 128.0 g of sulphur dioxide dissolved in every 1500 g of carbon dioxide?

3. A 123.0 g sample of hardwater is found to contain 0.015 g of calcium carbonate (CaCO3). What is it's concentration in ppm?

Answer 1

1. Molality= moles/kg

Moles= 18/180= 0.1

Molality= 0.1/1kg

Molality= 0.1mol/kg

2.SO2 emissions that lead to high concentrations of SO2 in the air generally also lead to the formation of other sulfur oxides (SOx). SOx can react with other compounds in the atmosphere to form small particles. These particles contribute to particulate matter (PM) pollution. Small particles may penetrate deeply into the lungs and in sufficient quantity can contribute to health problems.

3.Concentration is defined as moles/liter. So 0.012g of CaCO3 in 100.0g of water is 0.012g CaCO3/100.09g/mole / 100.0g water or 100.0ml or 0.100liters = 0.0012M CaCO3