Question

1. Read am authoritative source for a dicussion of filtration

2. What safety precautions should be taken when evaporating the filtrate containing NaCl?

3. Define the following terms as they apply to this experiment

(1) homogeneous mixture

(2) evaporation

(3) residue

4. A student was asked to determine the percentage of each each component in a mixture of sodium nitrate (NaNO3) and calcium carbonate(CaCO3). The mass of the sample used was 2.21g. The student extracted NaNO3 from the mixture with water and separated the insoluble CaCO3 from the soltuion by filtration. After evaporating the filtrate, the student recovered and dried the NaNO3, and found that it weighed .43g . The student dried the insoluble resiude of CaCO3 and found that it weighed 1.67g

(1) On the basis of the mass of sample used

(a) calculate percent NaNO3 in the mixture

(b) calculate percent CaCO3 in the mixture

(2) Calculate the total mass of the NaNO3 and CaC03 recovered

(3) Calculate the percent recovery of the compents, using the total mass of the substances recovered.

(4) Calculate the percent error for the separation of the components of the mixture

Answer 1

1. Filtration is a technique by which a solid is separated from solution called filtrate. Filtration is done using filter paper.

2. Safety precautions while doing filtration of NaCl solution would be, to wear proper safety wear as lab coat, safety glases, gloves and filtration done is a fume hood to avoid any spillage outside.

3. Definition

(1) Homogeneous solution - A single phase system wherein, all the components are soluble in one phase.

(2) evaporation : is a technique to convert a liquid phase system to a gaseous phase system.

(3) residue : The solid left on the filter paper after the filtration is done.

4. NaNO3 and CaCO3

(1) mass of NaNO3 collected = 0.43 g

mass of CaCO3 collected = 1.67 g

(a) percent NaNO3 in sample = 0.43 g x 100/2.21 g = 19.46%

(b) percent CaCO3 in sample = 1.67 g x 100/2.21 g = 75.56%

(2) total mass recovered = 0.43 + 1.67 = 2.1 g

(3) percent recovery of components = 2.1 g x 100/2.21 g = 95.02%

(4) percent error for separation = 100 - 95.02 = 4.98%