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fill in the blanks 3. Concentrations in solution. How many ml in 1 L? Ifyou diss

ID: 1043818 • Letter: F

Question

fill in the blanks

3. Concentrations in solution. How many ml in 1 L? Ifyou dissolve 25 g of a chemical in 100 ml, what is the concentration of the solution in g per L? concentration in solution in percent means grams per 100 mL. To make a 5.0% solution of a compound, how many gwould you weight out ifthe final volume of solution was 250 m? 4. Concentrations in solution. MolarityMnumber of mol per Ifthe molecularweight ofa compoundis 240gper mol, how many g would you weigh out to make 250 ml of a 1.0 M solution? If500 mL of a 0.1 M solution contains 20 g of a compound, what is its molecular weight? 5. Dilutions. If you take 25 m of a 500 ppb solution and add sufficient water until the volume is 100 m, and mix thoroughly, what is the concentration of the final solution? You dilute 10.0 mLof a 2.50 M solution to 500 ml, what is the final concentration?

Explanation / Answer

Ans. #3. 100.0 mL = 0.100 L

Now,

            [Chemical] = Mass of solute in g / Volume of soln. in liters

                                    = 25.0 g / 0.100 L

                                    = 250.0 g/ L

# Required mass = (% Concertation x Volume of solution) x 100

                                    = (5.0 % x 250.0 mL) x 100

                                    = [(5.0 g / 100.0 L) x 250.0 mL] x 100

                                    = 12.5 g

#4. Required moles = Molarity x Vol. of soln. in liters

                                    = 1.0 M x 0.250 L = 0.250 mol

Now,

            Required mass = Moles x MW = 0.250 mol x (240 g/ mol) = 60.0 g

# Moles of solute = Molarity x Vol. of soln. in liters

                                    = 0.1 M x 0.500 L = 0.050 mol

Now,

            MW = Mass of solute / Moles of solute = 20.0 g / (0.050 mol) = 400.0 g/ mol

#5. Using        C1V1 (original soln.) = C2V2 (diluted soln.)

            Or, C2 = (500 ppb x 25.0 mL) / 100.0 mL = 125 ppb

Therefore, final concertation = 125 ppb

# C2 = (2.50 M x 10.0 mL) / 500.0 mL = 0.05 M