Table 1 Salt Mass of salt, g Volume of HO, mL Temp (C) .qa79 q. S Density, gmL &

Question

Table 1 Salt Mass of salt, g Volume of HO, mL Temp (C) .qa79 q. S Density, gmL &.7°C 16.9 16-7 -49 F-06 3 5. 0019 04 5.ol0 50 KNO 20.1 11.20 28'. r.is 2. Calculate the solubility of the 5 salts in g solute per 100 g of water. Use the correct numboe of significant figures. If the salt did not dissolve in less than 50 g of water, report as insoluble. Report the temperature with the results. Put results in the form of a table. Assume the density of water is 1 g/mL so the volume and mass have the same value. 3. Calculate the solubility in molarity. Categorize the 5 salts as soluble or insoluble. Do your results agree with the solubility guidelines found in the text? 4. Include a second table in your report to present the results of questions 1 and 2. Table 2 Salt Naci KO Solubility (g solute per 100 g of water) Molarity (M) Temp (C) Cu50 Discussion Hints: (a) Order the salts from least soluble to most soluble. (b) Explain why the temperature is reported with the calculated solubility (c) Explain why the salts are ground into a fine powder with a mortar and pestle before dissolving

Explanation / Answer

2)

1) 4.997 g of NaCl in 16.7 mL H2O

Therefore, solubility in 100 g water (Density of water =1, therefore, 100 mL water = 100 g water).

= (4.997/16.7) x 100 = 29.92 g

2) 5.001 g of KNO3 in 44.5 mL H2O

Therefore, solubility in 100 g water = (5.001/44.5) x 100 = 11.23 g

3) 5.010 g of CaSO4 in 50 mL H2O

Therefore, solubility in 100 g water = (5.010/50) x 100 = 10.02 g

4) 5.017 g of CuSO4 in 28.4 mL H2O

Therefore, solubility in 100 g water = (5.017/28.4) x 100 = 17.66 g

5) 4.998 g of NH4NO3 in 6.5 mL H2O

Therefore, solubility in 100 g water = (4.998/6.5) x 100 = 76.89 g

Solubility order is,

CaSO4 < KNO3 < CuSO4 < NH4NO3

Solubility varies with temperature, therefore, temperature is reported along with the solubility data

Salts are usually ground in to fine powder to increase the surface area. With increase in surface area, the contact area with the solvent molecule increases and result in easy dissolution

3)

a) Molarity of NaCl solution

Molar mass of NaCl = 58.4 g/mol

Number of moles of NaCl = (29.92 /58.4 g/mol) = 0.512

Therefore, molarity = Number of moles of NaCl/Liters of solution = 0.512 /0.1 = 5.12 M

b) Molarity of KNO3 solution

Molar massof KNO3 = 101.103 g/mol

Number of moles of KNO3 = (11.23 /101.103 g/mol) = 0.111

Therefore, molarity = Number of moles of NaCl/Liters of solution = 0.111 /0.1 = 1.11 M

c) Molarity of CaSO4 solution

Molar mass of CaSO4 = 136.14 g/mol

Number of moles of CaSO4 = (10.02 /136.14 g/mol) = 0.073

Therefore, molarity = Number of moles of NaCl/Liters of solution = 0.073 /0.1 = 0.73 M

d) Molarity of CuSO4 solution

Molar mass of CuSO4 = 159.609 g/mol

Number of moles of CuSO4 = (17.66 /159.609 g/mol) = 0.110

Therefore, molarity = Number of moles of NaCl/Liters of solution = 0.110 /0.1 = 1.1 M

e) Molarity of NH4NO3 solution

Molar mass of NH4NO3 = 80.043 g/mol

Number of moles of NH4NO3 = (76.89 /80.043 g/mol) = 0.960

Therefore, molarity = Number of moles of NaCl/Liters of solution = 0.960 /0.1 = 9.6 M

Salt Solubility (g solute per 100 g of water) Molarity (M) Temp (C ) NaCl 29.92 g 5.12 M 18.7 KNO3 11.23 g 1.11 M 16.9 CaSO4 10.02 g 0.73 M CuSO4 17.66 g 1.1 M 20.1 NH4NO3 76.89 g 9.6 M 11.3

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