a. Calculate the mass (in grams) of potassium hydroxide in 500 cm^3 of a KOH-H2O

Question

a. Calculate the mass (in grams) of potassium hydroxide in 500 cm^3 of a KOH-H2O solution at a concentration of 2 M KOH solution. b. A mixture of gases has the following composition by mass: O2 10.0% CO 10.0% CO2 15.0 % N265.0 % Determine the molar composition and the average molecular weight of the gas mixture. c. A hydrocarbon contains 85.714 mass% carbon (C) and 14.286 mass % hydrogen (H) Its molecular weight is 42. Determine its chemical formula (i.e. what would be x and y in CxHy). d. Magnesium sulphate crystals (MgSO4.nH2O) have 51.2 mass% water and 48.8 mass% MgSO4. Determine n in the formula MgSO4.nH2O. How to solve this plz

Explanation / Answer

a) 1000 cc = 1 litres

Now, 500 cc = 0.5 litres

Now, moles of KOH in the solution = molarity*volume of solution in litres = 2*0.5 = 1

Molar mass of KOH = 56 g/mole

Thus, mass of KOH = moles*molar mass = 56 g

b) Let the mass of the mixture be 100 g

Now, mass of O2 = 10 g

mass of CO = 10 g

Mass of CO2 = 15 g

Mass of N2 = 65 g

Now, molar mass of O2= 32 g/mole ; CO = 28 g/mole ; CO2 = 44 g/mole & N2 = 28 g/mole

Thus, moles of 10 g of O2 = 10/32 = 0.3125

moles of CO = 10/28 = 0.357

moles of CO2 = 15/44 = 0.341

moles of N2 = 65/28 = 2.32

Thus, total moles of gases present = 6.143

Thus, average molecular weight of the gaseous mixture = mass of the gaseous mixture/moles of the gases in total = 100/6.143 = 16.28 g/mole

c) Let the formula of the hydrocarbon be CxHy

Let the mass of the hydrocarbon be 100 g

Now, mass of C = 85.714 g & mass of H = 14.286 g

Now, molar mass of C = 12 g/mole & molar mass of H = 1 g/mole

Thus, moles of C present = 85.714/12 = 7.143

moles of H present = 14.286

Thus, ratio of the moles of C & H in 100 g of the compound = 7.143:14.286 = 1:2

Thus, the empirical formula of the compound = CH2

Now, let the molecular formula be (CH2)n

Thus, molar mass = 14*n

Now, given molar mass = 42

Thus, 14n = 42

or, n = 3

Thus, the molecular formula is :-

C3H6

d) Molar mass of MgSO4.nH2O = 120 + 18*n

Now, mass of MgSO4 in 1 mole of the compound = 120 g

Thus, mass % of MgSO4 in the compound = (mass of MgSO4/mass of the compound)*100 = (120/120+18n)*100

or, 120/(120+18n) = 0.488

or, n = 7

  

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