2 100.0 mL of 2.500 M KBr solution is on hand. You need 0.5500 M. What is the fi

Question

2 100.0 mL of 2.500 M KBr solution is on hand. You need 0.5500 M. What is the final 3.) 15.0 mL of 0.309 M Na,SO, and 35.6 mL of o.200 M KCI are mixed. Determine the 4. How many milliliters of 1.5 M AICl, must be used to make 70.0 mL of a solution tha 8) A 40% acid solution is mixed with a 75% acid solution to produce 140 liters of a 509 6. Given a density of 1.769 g/ml, and a H.SO, mole fraction of 0.5000, find the volume of solution which results? following concentrations: [Na']; [SO2: [Cl] has a concentration of 0.21 M cI? acid solution? How many liters of each acid solution was mixed? molality, molarity, and mass percentas p what is the molarity of a 30.0% (w/w) hydrogen peroxide solution? Reagent grade nitric acid is 70.40% HNO3 (63.0119 g/mol) by mass and its molarity is 16.00 M. Calculate the density, molality and mole fraction of nitric acid in the solution. 7, 8. 9. The density of toluene (C,Hs) is 0.867 g/mL, and the density of thiophene (C4HH4S) is 1.065 g/mL. A solution is made by dissolving 9.660 g of thiophene in 260.0 mL toluene.solvn a. Calculate the molality of thiophene in the solution. b. Assuming that the volumes of the solute and solvent are additive, determine 10. The vinegar sold in the grocery stores is described as 5% (v/v) acetic acid, what is 11. By titration, the molarity of acetic acid in vinegar was determined to be 0.870 M the molarity of thiophene in the solution. the molarity of this solution (density of 100% acetic acid is 1.05 g/mL)? Convert this to %(v/v). (The density of acetic acid is 1.05 g/mL)

Explanation / Answer

2.

M1 V1 = M2 V2

Or, (2.5 M) x (100 mL) = (0.55 M) X V2

Or, V2 = (2.5 M) x (100 mL) / (0.55 M)

Or, V2 = 454.55 mL

4.

M1 V1 = M2 V2

Or, (1.5 M) x V1 = (0.21 M) X (70 mL)

Or, V1 = (0.21 M) X (70 mL) / (1.5 M)

Or, V2 = 9.8mL

7.

30%(w/w) means

30 g of H2O2 in 100 g of solution
or, 300 g of H2O2 in 1000 g of solution
or, 300 g of H2O2 in 1000 mL of solution           (density of water = 1 g/mL)
or, 300 g of H2O2 in 1 L of solution.

Molar mass of H2O2 = 34 g/mol

So, 34 g of H2O2 = 1 mol
or, 1 g of H2O2 = (1/34) mol
or, 300 g of H2O2 = (300/34) mol = 8.82 mol

So, 8.82 mol of H2O2 in 1L, which means = 8.82 M

11.

0.870 M solution means

0.870 moles of acetic acid in 1000 mL vinegar solution

Molar mass of acetic acid = 60 g/mol
So, 1 mole of acetic acid = 60 g
or, 0.870 mole of acetic acid = 0.870 x 60 g = 52.2 g

Mass of acetic acid = 52.2 g
Density of acetic acid = 1.05 g/mL
So, volume of acetic acid = Mass / Volume
                                              = (52.2 g) / (1.05 g/mL)

                                             = 49.71 mL

So, 49.71 mL in 1000 mL

%(v/v) = (49.71 mL / 1000 mL) x 100
            = 4.97 %

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