EXPERIMENT6 COMPOUND DETERMIGNG THE SOLUBILITY OF AN REPORT Post-Lob Questions U

Question

EXPERIMENT6 COMPOUND DETERMIGNG THE SOLUBILITY OF AN REPORT Post-Lob Questions UNKNOWN IONIC NAME DATE of ionie 1. Using either your lecture's texthook or wour unknown compounds), what is a resosurce (search solubility eurves n possible identity of your unknown compound? identity 2. In the BACKGROUND information, it ws tion, it was stated that CaFs has solubility, at room tempe 00 e'ml how many moles of CaFi will dissolve in exactly 1.00 of 0.00160 g per 100 g of water. C ms of solution? n? tion of CaF, is ow many moles of CaF is this? Assuming that the density of CaFs is 1. Using the solubility curve for maximum amount of grams of 250.0 mL of water at 55 C? 3. KCr 0 to the right, what is the0 70 KgCro07that would dissolve in 60 50 40 30 20 10 o 10 20 30 40 50 60 70 80 90 100 Temperature (C) 110 EXPERIMENT 6

Explanation / Answer

Ans 2

Solubility of CaF2 at room temperature

= 0.00160g / 100g water

Moles of CaF2 = mass/molecular weight

= 0.00160g / 78.07 g/mol

= 2.049 x 10^-5 mol

Total mass of solution = 0.00160 + 100 = 100.00160 g

Density of solution = 1 g/mL

Volume of solution = mass/density

= 100.00160 g / 1 g/mL

= 100.00160 mL x 1L/1000 mL

= 0.1000016 L

0.1000016 L solution dissolves = 0.00160g CaF2

1 L solution will dissolve = 0.00160 g CaF2 /0.1000016 L

= 0.0159997 g / L

Moles of CaF2 dissolved in 1 L solution

= 0.0159997 g / 78.07 g/mol

= 0.00020494100 mol

= 2.049 x 10^-4

Ans 3

From the graph

At 55°C

Solubility of K2Cr2O7 = 34g / 100gH2O

Density of water = 0.98569 g/mL

Mass of water = 0.98569 g/mL x 250 mL

= 246.4225 g

Maximum amount of K2Cr2O7

= (34g / 100 g H2O) x (246.4225 g H2O)

= 83.78365 g K2Cr2O7

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