Inorganic questions--please show work and only answer if you truly understand it
ID: 504864 • Letter: I
Question
Inorganic questions--please show work and only answer if you truly understand it:
Please show all work and answers on this page if at all possible.
6. The alcohol content of hard liquor is normally given in terms of the “proof”, which is defined as twice the percentage by volume of ethanol (C2H5OH) present. Calculate the number of grams of alcohol present in a glass of of 75-proof gin (assume a 4 oz glass). The density of ethanol is 0.798 g/mL.
7. You find a solution on the shelf that is labeled 30% NH3 (aq). The density of the solution is 0.982 g/mL. Calculate the following:
a) mole fraction of ammonia = ____________
b) molality of ammonia = ____________
c) molarity of ammonia = ____________
8. Adrenaline is the hormone that triggers release of extra glucose molecules in times of stress or emergency. A solution of 0.64 g of adrenaline in 36.0 g of CCl4causes an elevation of 0.49C in the boiling point. Given that the kb for CCl4 is 5.02 C/m, what is the molecular mass of adrenaline.
9. The Henry's Law constant for the solubility of nitrogen in water is 6.40x10-4M/atm at 25.0 Celsius. At 570 torr, how many grams of N2 can be dissolved in 250 L of water at 25.0 degrees Celsius
10. What is the osmotic pressure of a 1.36 M aqueous solution of urea [(NH2)2CO)] at 22.0C ?
Explanation / Answer
Q6. Gin is 75-proof hence percentage of ethanol by volume = 75 / 2 = 37.5%V/ V
It implies 100mL of gin contain 37.5mL of ethanol (C2H5OH).
Volume of gin = 1 glass = 4 oz = 4 X 29.57mL = 118.28mL
Volume of ethanol in gin = 118.28 X 37.5 /100 = 44.35mL
Mass of ethanol = Density X Volume = 0.798g/mL X 44.35mL = 35.39g
Q6. Lets consider 100mL of the solution
Mass of solution = density X Volume = 0.982g/mL X 100 = 98.2g
Mass of ammonia = Mass of solution X Percentage of ammonia = 98.2g X 30 / 100 = 29.46g
Hence mass of water = 98.2g - 29.46g = 68.74g
Moles of NH3 = Mass of NH3 / Molae mass = 29.46g / 17 = 1.7 moles
Moles of water = 68.74g / 18g = 3.82 moles
(A) Mole fraction of ammonia = moles of ammonia / total moles = 1.7 / (1.7 + 3.82) = 0.31
(B) Molality of ammonia = (moles of ammonia / mass of water) X 1000 = (1.7 / 68.74 ) X 1000 = 24.73m
(C) Molarity of ammonia = (Moles of ammonia / Volume of solution in mL ) X 1000
= (1.7 / 100mL) X 1000 = 17 M
Because 100mL solution contain 1.7 moles of ammonia
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