John has to prepare two solutions: Solution A: 500 mL of 0.2 M solution of potas

Question

John has to prepare two solutions: Solution A: 500 mL of 0.2 M solution of potassium bronside (KBr) in water Solution B: 200 mL of 0.4 M solution in of glucose in water How much of KBrand how much of glucose (in g) John have to take to prepare these two solutions? John mixed 50 mL of solution A and 150 mL of solution B. What are concentrations (in M) of KBr and glucose in the mixture? MAry has to prepare 250 mL of 1 M solution of sulfuric acid. How much (in mL) of commercially available concentrated sulfuric acid (concentration 18 M) she should take to prepare this solution? Mary added 20 mL of pure (100%) ethanol to the 1 L of water. What is (approximately) molor concentartion of alcohol in the resulting solution (density of ethanol is 0.789 g/cm3)? Mary added 3 g of potassium permanganate (KMnO4) to 47 mL of water. What is % concentration of the resulting solution? Jphn has to prepare 1 kg of 2% solution of sodium chloride (NaCl). How much NaCl (in g) and how much of

Explanation / Answer

1.a)Solution A - 500ml of 0.2 M KBr

Moles of KBr = 0.5 L x 0.2 M = 0.1 moles

Solution B -200 mL of 0.4 M glucose

Moles of glucose= 0.2 L x 0.4 M = 0.08 moles

Grams of KBr that should be added = moles of KBr x molar mass of KBr

                                                     = 0.1 mole x 119.002 g/mol =11.9 g

Grams of glucose that should be added = moles of glucose x molar mass of glucose

                                                           = 0.08 moles x 180 g/mol = 14.4 g

b)

Total volume = 50 + 150 = 200 mL = 0.2 L

Concentration of KBr = 0.1/0.2 = 0.5 M

Concentration of glucose = 0.08 moles/0.2 L = 0.4 M

2.

M1V1 = M2V2

250mL x 1 M = 18 M x V2

V2 = 250 x 1/ 18 = 13.88 mL

3.

Density = 0.789g/cm3

Mass = density x volume = 0.789 g/mlx 20 ml= 15.78 g

Moles of ethanol = given mass/molar mass = 15.78g/46.06 g/mol = 0.342 moles

Molar concentration = moles in litres of solution = 0.342/1L = 0.342 M

4.a)

Moles of KMnO4 = mass of KMnO4/Molar mass of KMnO4 = 3g/158 g/mol = 0.0189 moles

Concentration= moles of solute/litres of solution = 0.0189moles/0.047 L = 0.402 M

b)

1 Kg of 2% solution

That means total mass 1000 g

NaCl = 20 g

water = 980 g water

Volume of water = mass/density = 980g/1.00 g/ml= 980 ml

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