Basic Chemistry

Question

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1. 7.50 g of sodium chloride are added to 3200 mL of water. Find the mole fraction of the sodium chloride and of the water in the solution. (note: 1 mole = 22.4 L)
2. How many grams of magnesium cyanide (MgCN2) are needed to make 300 mL of a 0.095 M solution?
3. What is the molality of 75.0 g of glucose (C6H12O6) dissolved in 125 g of water?
4. To make a 65% methyl alcohol solution by volume. What volume of methyl alcohol should be added to 150 mL of water to make this solution?

Answer 1

1). Mole fraction is moles / total moles

moles NaCl = 7.50 g NaCl x 1 mol NaCl / 58.4 g = 0.128 mols NaCl

moles H2O = 3200 mls x 1 g / ml = 3200 g x 1 mol / 18 g = 178 mols H2O

Total moles = 0.128 + 178 = 178.128 moles

Mole fraction NaCl = 0.128 / 178.128

Mole fraction H2O = 178 / 178.128

The fact that 1 mole = 22.4 L only applies to gases at STP. It has nothing to do with this problem. And this problem is ridiculous based on the values given. The mol fraction of H2O is essentially 1.

2). Molar mass Mg(CN)2 = 76.3 g / mole (NOTE: it is NOT MgCN2

0.095 mols / L x 0.300 L x 76.3 g / mole = 2.17 g needed

3). molality = moles of glucose per kg of water

Molar mass glucose = 180 g / mol

75.0 g x 1 mol / 180 g = 0.417 mols glucose

125 g H2O x 1 kg / 1000 g = 0.125 kg H2O

m = 0.417 mols / 0.125 kg = 3.34 m

4). 65% = 0.65

x ml methyl alcohol / 150 ml + x ml = 0.65

97.5 + 0.65x = x

0.35x = 97.5

x = 279 mls of methyl alcohol should be added to 150 mls