44 gof C molarity, M, of the solution? Show your work. dissolved in enough water
ID: 1043773 • Letter: 4
Question
44 gof C molarity, M, of the solution? Show your work. dissolved in enough water to make 500 mL of solution, what will be the molCaCh 500 3) A sample vial containing 1.25 g of compound AB weighs 18.46g (including AB). After adding small portions of water, the compound finally dissolves. The vial and contents now weigh 33.56 g. Calculate the solubility of AB in the following units. Show your work. w/w % (ie, g AB/100g solution ) ppm 4) Give the names and formulas for the compounds listed in Table 1 of this experiment. In 442 there are handbooks you can use to find formulas for those that are listed by name.Explanation / Answer
ANSWER:
Molarity (M) of a solution is given as = no. of moles of solute dissolved / volume of solution in Liters
Moles of CaCl2 = Given mass of CaCl2 / molar mass of CaCl2 = 44g / 110.98g = 0.396 moles
Volume = 500mL = 500mL / 1000 = 0.5L
Threfore , M = 0.396 moles / 0.5 L= 0.79M
Q2. The solubility in %(W/W) is given as
% (W/W) = (mass of solute / mass solution) X 100
Mass of solute = 1.25g
Mass of solution = final mass of vial - initial mass of vial = 33.56g - 18.46g = 15.1g
% (W/W) = (1.25 / 15.1) X 100 = 8.28 %
ppm concentration of the solution = (1.25g / 15.1 g ) X 106 = 82781.46ppm
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