Solution | Questions based on Concentration Terms
Questions
Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution containing 20% of C2H6O2 by mass.
Let the mass of solution be 100 grams
Then, the mass of C2H6O2 = 20 grams.
And, mass of water = (100 - 20) grams = 80 grams.
Molecular mass of C2H6O2 = 62 grams/mole.
Moles of C2H6O2 = 20/62
Moles of water = 80/18
\[Χ_{C_2H_6O_2} = {n_{C_2H_6O_2} \over n_{C_2H_6O_2} + n_{H_2O}}\]
\[Χ_{C_2H_6O_2} = {20/62 \over 20/62 + 80/18} = 0.068\]
Also,
\[Χ_{H_2O} = 1 - 0.068 = 0.932\]
Calculate the molarity of solution containing 5 g of NaOH in 450 mL solution.
\[n_{NAOH} = {5 \over 40} = 0.125\ moles\]
\[V_{solution} = {450 \over 1000} = 0.45\ L\]
\[M = {n_{NAOH} \over V_{solution}} = {0.125 \over 0.45} = 0.278\ mol/L\]
Calculate the molality of 2.5 g of ethanoic acid (CH3COOH) dissolved in 75 g of benzene (C6H6).
\[Molecular\ mass\ of\ CH_3COOH = 60\ g/mol\]
\[n_{CH_3COOH} = {2.5 \over 60} = 0.0417\ moles\]
\[m_{solvent} = {75 \over 1000} = 0.075\ Kg\]
\[m = {n_{CH_3COOH} \over m_{solvent}}\]
\[m = {0.0417 \over 0.075} = 0.556\ mol/Kg\]
Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
\[Mass\ of\ solution = 22 + 122 = 144\ g\]
\[w/w\ of\ benzene = {22\times 100\over 144} = 15.28\]
\[w/w\ of\ CCl_4 = {122 \times 100\over 144} = 84.72\]
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Let the mass of solution be 100 g
Then, the mass of benzene = 30 g
And, the mass of carbon tetrachloride = 70 g
Moles of benzene (nC6H6) = 30/78 g/mol
Moles of carbon tetrachloride (nCCl4) = 70/154 g/mol
\[Χ_{C_6H_6} = {n_{C_6H_6} \over n_{C_6H_6} + n_{CCl_4}}\]
\[Χ_{C_6H_6} = {30/78 \over 30/78 + 70/154} = 0.459\]
\[Χ_{CCl_4} = 1 - 0.459 = 0.541\]
Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2.6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.
(a)
\[Moles\ of\ Co(NO_3)_2.6H_2O = {30 \over 291}\]
\[Molarity = {30/291 \over 4.3} = 0.024 mol/L\]
(b) Here the moles of solute will be same even after dilution.
\[n_{H_2SO_4} = M_{H_2SO_4} \times V_{solution} (L)\]
\[n_{H_2SO_4} = {0.5 \times 0.3} = 0.15\]
Molarity of diluted solution is given as:
\[M = {n_{H_2SO_4} \over V_{solution} (L)}\]
\[M = {0.15 \over 0.5} = 0.03\ mol/L\]
Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.
\[Let\ mass\ of\ urea = x\ kg\]
\[Molality = 0.25\ m\]
\[Mass of solution = 2.5\ kg\]
\[Mass of solvent = (2.5 - x)\ kg\]
\[Moles of urea = {x\times 1000\over 60}\]
\[Molality = {(x\times 1000)/60\over 2.5-x}\]
\[=>\ 0.25 = {(x\times 1000)/60\over 2.5-x}\]
\[=>\ x = 0.034946\ kg = 34.946\ g\]
Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g/mL.
\[w/w = 20\]
\[The\ solvent\ is\ water.\]
\[Density\ of\ solution = 1.202\ g/mL\]
\[Let\ mass\ of\ solution\ be\ 100\ g\]
\[Then,\ mass\ of\ KI = 20\ g\]
\[And,\ mass\ of\ solvent = 80\ g\]
\[Moles\ of\ solvent = {80\over 18}\]
\[Volume\ of\ solution = {100\over1.202} mL\]
\[Moles of KI = {20\over166}\]
\[molality = {20/166 \over 80/1000} = 1.506\ mol/kg\]
\[molarity = {20/166 \over 100/(1.202\times 1000)} = 1.45 mol/L \]
\[Χ = {20/166\over (20/166 + 80/18)} = 0.0263\]