Specific conductance is a measure of water\'s capacity to conduct an electric cu

Question

Specific conductance is a measure of water's capacity to conduct an electric current, which is determined by the concentration concentration 4. of dissolved ions (electrolytes)-- the greater the of dissolved ions, the higher the conductivity. Resistance is a measure of the through a material. Resistance is measured in difficulty electrons or ions encounter moving through a material Resistance is meas ohms. Conductivity, measured in Siemens, is the inverse of resistance. In general the relationship between conductivity and ion concentration is linear. Conductivity measurements are commonly used as a quality control check of labotatory reagents to determine if they have been made properly. It also can be used clinically to indicate electrolyte levels in body fhuids. There are many other applications. Have the instructor measure the specific conductance of your 1.6 x 10'M NaCl solution and record this value on Table 4. Complete the calculations. 5. Project 2: Making a percent composition solution In this project you will use calculations, conversions, and dilutions to make a 0.014% solution. 1· Clean out your volumetric flask from Project 1 and make 100 mL of a 3.5% solution of NaCl. 2. Make a-dilution of the 3.5% NaCl stock solution by adding 1 mL of the 3.5% solution to 9.0 10 mL of water in a test tube. Label this test tube as 035%. Seal the test tube with a piece of paraflm and invert the test tube several times to mix the contents. dit on of 3.5% Naa-in x 3.5% = 0.35% NaCl --ig/ml. 10 10 Make a 1 /25 dilution of the 0.35% NaCl solution by placing 2 mL of the 0.35% NaCl solution into a 100-mL pour beaker and use a 50-mL graduated cylinder to add 48 mL of H,O to the beaker. Add a parafilm seal and swirl the beaker to mix the contents. 3. What is the concentration of this solution % NaCl; ug/mL Have the instructor measure the specific conductance of your final dilution and record this valuc on Table 1. Calculate the percent deviation for your sample. 4.

Explanation / Answer

Project 2

(2) 3.5% NaCl solution here is 3.5 gm in 100ml water

1/10 th dilution of 3.5% NaCl solution= 1/10 * 3.5% = 0.35 % NaCl solution = 0.35gm in 100ml water

= 350 mg NaCl in 100 ml water

= 3.5mg NaCl in 1 ml water

= 3500ug in 1ml water

= 3500mg/ mL water

(3) 1/25 * 0.35 % = 0.014 % NaCl solution

      = 0.014 gm NaCl in 100 ml water

= 14000ug NaCl in 100 ml water

= 140 ug/ mL water

P.S.- 1gm = 1000,000 ug

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