I need help in Chemistry...

Question

What is the step-by-step procedure and solution on solving these?

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?

Answer 1

1). They are asking you to find the mole fraction of the NaCl and H2O. The mole fraction of something is the number of moles of that compound divided by the total number of moles present. (The "note" that 1 mole = 22.4 L has nothing to do with this problem as that value relates to ideal gases). So here are the steps:

-- find moles of NaCl and moles of H2O:

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

moles H2O = 3200 ml H2O x 1 g / ml x 1 mol H2O / 18 g = 178 mols H2O (assuming a density of H2O = 1g/ml

-- find total moles:

total moles = 0.128 mols + 178 mols = 178.128 mols

-- find mole fraction of each:

mole fraction NaCl = 0.128 mol / 178.128 mol = 0.000719

mole fraction H2O = 178 / 178.128 = 0.999

2), The definition of 0.095 M Mg(CN)₂ is 0.095 mols Mg(CN)₂ per liter of solution.

step1- determine moles of Mg(CN)₂ needed to make 300 mls:

300 mls x 1 L /1000 mls x 0.095 mols / L = 0.0285 mols Mg(CN)₂ needed to make 300 mls

step 2 - determine grams Mg(CN)₂ needed using the molar mass of Mg(CN)₂ = 76.3 g/mol:

0.0285 mols x 76.3 g / mol = 2.17 g Mg(CN)₂ needed

3). The definition of molality is moles of solute (glucose) per kg of solvent (water).

step 1 - find moles of solute using molar mass of glucose = 180 g/mol:

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

step 2 - find kg of water:

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

step 3 - divide moles by kg:

molality = 0.417 mols / 0.125 kg = 3.33 m