##### Number of moles present

 Chemistry Tutor: None Selected Time limit: 1 Day

Calculate the number of moles of solute present in each of the following aqueous solutions.

(a) 600. mL of 0.268 M SrBr2
mol

(b) 84.3 g of 0.175 m KCl
mol

(c) 132.0 g of a solution that is 6.30% glucose, C6H12O6, by mass.
mol
Mar 13th, 2015

a) The volume of solution is 600 mL = 0.600 L (consider with 3 significant figures.

Molarity is 0.268 mol/L. The amount of SrBr2 is 0.600*0.268 = 0.161 mol.

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

The mass of the solute is 0.175 mol * (39.10 + 35.45) g/mol = 0.175 * 74.55 = 13.0 g and the total mass of solution is 1 kg + 13.0 g = 1.013 kg.

Write the proportion 0.175 mol / 1.013 kg = x mol / 0.0843 kg.

Then x = 0.175*0.0843/1.013 = 0.0146 mol of KCl.

c) The mass of glucose in the solutions is 6.30% of 132.0 g = 8.316 g.

The molar mass of glucose is 6*12.011 + 12*1.008 + 6*16 = 180.2 g/mol.

The number of moles is 8.316 / 180.2 = 0.0461 mol.

Mar 13th, 2015

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Mar 13th, 2015
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Mar 13th, 2015
Dec 10th, 2016
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