Describe how you would prepare 4.0 L a 0.154 M solution of NaCl using 1.0 M NaCl
ID: 612777 • Letter: D
Question
Describe how you would prepare 4.0 L a 0.154 M solution of NaCl using 1.0 M NaCl solutionExplanation / Answer
To prepare this solution you first need to work out how many moles of NaOH you need. Molarity = moles/litres Therefore moles = Molarity x litres You know that M = 0.20 litres = 2.00 so moles = 0.20M x 2.00 litres = 0.4 moles so you need 0.4 moles of NaOH to prepare 2.00 L of 0.20 M NaOH Now calculate the mass of 0.4 moles of NaOH moles = mass / molecular weight molecular weight of NaOH = 22.99 + 16.0 + 1.008 = 40.00 g/mol therefore mass = moles x molecular weight mass = 40.00 x 0.4 = 16.00 g To make the solution weigh 16.00 g of NaOH into a 2.00 L volumetric flask. Add a few hundred ml of water and dissolve the solid NaOH. Once all the solid has dissolved make the solution up to 2.00 L with water. It is important to dissolve all the solid first because some substances alter the volume of a solution on dissolving. NaCl - think about what is happening when water boils. Liquid water is turning to the gas phase thus being removed from the solution. The amount of NaCl in the solution is remaining the same. So the solution is becoming more concentrated because the same amount of salt is dissolved in less water Molarity = moles/litres 5) You are told that the salts are strong electrolytes. Strong electrolytes dissociate fully in solution, so you can assume that this is the case here. The first thing you need to do is write balance equations forthe dissolution. a) CaCl2 -------> Ca(2+)(aq) + 2 Cl(-)(aq) the balanced equation tells us that 1 mole of CaCl2 yields 1 mole of Ca(2+) and 2 moles of Cl(-) If the solution prepared was 0.25 M of CaCl2 then: the concentration of Ca(2+) will be 0.25 M (because 1 mole of salt gives one mole of Ca(2+)) the concentration of Cl(-) will be 0.5 M (because 1 mole of CaCl2 dissociates to give 2 moles of Cl(-)) b) Al(NO3)3 -------------> Al(3+)(aq) + 3NO3(-)(aq) Apply the same logic to this as to the question above. Al(3+) will be 0.15 M and NO3(-) will be 0.45 M c) K3PO4 -------------> 3K(+)(aq) + PO4(3-)(aq) For this question you need to work out the concentration of the solution first using the equations moles = mass / molecular weight Molarity = moles / litres molecular weight of K3PO4 = (9.1 x 3)+ 30.97 + (6 x 4) = 212.27 g/mol therefore moles of K3PO4 = mass / molecular wt = 42.4g / 212.27 g/mol =0.1997 moles of K3PO4 in the solution Now molarity = moles / litres M= 0.1997 mol/ 0.250 litre Molarity = 0.80 moles/litre Now, again apply the same logic as show in a) to work out the concentrations of the salts: K(+) will be 2.4 M and PO4(3-) will be 0.8M 1) a) the solute is the substance being dissolved (KNO3) b) the solvent is what the solute is being dissolved in (deionized water) c) this is the percentage of the total mass of the solution that the KNO3 comprises % = mass of KNO3 / total mass of solution x 100/1 % = (6 / (6 + 14)) x 100 / 1 = 30% m/m (mass/mass)