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Chemistry
Tutor: None Selected Time limit: 1 Day

Mar 13th, 2015

a) Volume of solution = 550 mL = 0.550 L; molarity = 0.217 M

The amount of the solute = 0.550 L * 0.217 mol/L = 0.119 mol

b) The molality of solution is 0.175 m. It means that 0.175 mol of KCl is dissolved in 1 kg of water. 

1 mol of KCl = 39.10 + 35.45 = 74.55 g of KCl. (use the periodic table for atomic masses of K and Cl)

Then 0.175 mol  contain  0.175 * 74.55 = 13.0 g of KCl.  The total mass of solution is 1.013 kg. 

From the proportion 0.175 mol / 1.013 kg = x mol / 0.0818 kg we find that x = 0.175 *0.0818/1.013 = 0.141 mol 

c) The mass of the solute (glucose) is 120.0 g * 6.75% = 8.1 g, however, we need to find the number of moles. Since the molar mass of C6H12O6 is 6*12.011+12*1.008+6*16 = 180 g/mol, the number of moles is 

8.1 g / 180 g/mol = 0.045 mol

Mar 13th, 2015

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