Chapter 1: Solutions

Ans: Mass percentage is the ratio of the mass of the solute divided by the total mass of the solution multiplied by 100. 

$Mass\%=\dfrac{mas{{s}_{solute}}}{mas{{s}_{solution}}}\times 100$ 

Here, 

The mass percentage of both compounds is given as;

Mass percentage of ${{C}_{6}}{{H}_{6}}$= $\dfrac{mas{{s}_{{{C}_{6}}{{H}_{6}}}}}{mas{{s}_{{{C}_{6}}{{H}_{6}}}}+mas{{s}_{CC{{l}_{4}}}}}\times 100$ 

                                        = $\dfrac{22}{22+122}\times 100$ 

                                        = 15.28 %

Now, as the solution consists of only two components i.e., ${{C}_{6}}{{H}_{6}}$ and $CC{{l}_{4}}$.

Thus,

Mass percentage of $CC{{l}_{4}}$= 100 – 15.28

                                        = 84.72%

Therefore, the mass percentages of ${{C}_{6}}{{H}_{6}}$ and $CC{{l}_{4}}$ are 15.28% and 84.72% respectively.

Ans: Mole fraction is the ratio of moles of solute to the total moles of solution.

Mole fraction = $\dfrac{{{n}_{solute}}}{{{n}_{solution}}}$ 

Here, 

We have given a solution containing 30% by mass of benzene in carbon tetrachloride; which clearly states that 30 g of benzene is present in 100 g of solution.

Thus, carbon tetrachloride is 70 g in 100 g of solution.

We know that,

Molar mass of benzene = 78 g/mol

Molar mass of carbon tetrachloride = 154 g/mol

Thus,

Number of moles of benzene = $\dfrac{30}{78}=0.385mol$  

Number of moles of carbon tetrachloride = $\dfrac{70}{154}=0.455mol$ 

Now, 

Mole fraction of benzene = $\dfrac{{{n}_{{{C}_{6}}{{H}_{6}}}}}{{{n}_{{{C}_{6}}{{H}_{6}}}}+{{n}_{CC{{l}_{4}}}}}$ 

                                         = $\dfrac{0.385}{0.385+0.455}$ 

                                         = $\dfrac{0.385}{0.84}=0.458$ 

Therefore, the mole fraction of benzene is 0.458 in the solution.

Ans: Molarity is the number of moles of solute per liter of solution.

Molarity = $\dfrac{{{n}_{solute}}}{{{V}_{solution}}}$ M

Here,

Molarity of 30 g of $Co{{\left( N{{O}_{3}} \right)}_{2}}.6{{H}_{2}}O$  in 4.3 L of      

the solution is given as;

Molar mass of $Co{{\left( N{{O}_{3}} \right)}_{2}}.6{{H}_{2}}O$ = 310.7 g/mol.

Number of moles = $\dfrac{30}{310.7}=0.0966$ 

Thus, molarity = $\dfrac{0.0966}{4.3}=0.022M$

1. 30 mL of 0.5 M ${{H}_{2}}S{{O}_{4}}$  diluted to 500 ml.

Ans: Molarity of 30 mL of 0.5 M ${{H}_{2}}S{{O}_{4}}$  diluted to 500 ml.

1000 ml of 0.5 M ${{H}_{2}}S{{O}_{4}}$ contains 0.5 moles of ${{H}_{2}}S{{O}_{4}}$.

Thus,

30 ml of 0.5 M ${{H}_{2}}S{{O}_{4}}$ contains;

= $\dfrac{0.5}{1000}\times 30=0.015moles$ ${{H}_{2}}S{{O}_{4}}$.

Also, the volume of solution = 500 ml = 0.5 L.

Now,

Therefore the molarity = $\dfrac{0.015}{0.5}=0.03M$

Ans: Molality is given as ratio of moles of solute per gm of mass of solvent.

Molality = $\dfrac{{{n}_{solute}}}{{{m}_{solvent}}}$ m

Here,

0.25 molal solution of urea states that 0.25 moles of urea is present in 1000 gm of solvent.

The molar mass of solute i.e., urea = 60 g/mol

Thus,

Mass of urea in the solution = $0.25\times 60=15g$ 

Total mass of solution = 1000 gm solvent + 15 gm of solute 

                                     = 1015 gm = 1.015 Kg

This states that, 

1.015 Kg solution has 15 gm urea.

Thus,

2.5 Kg solution will have;

$\dfrac{15}{1.015}\times 2.5=37g$ urea.

Therefore, the mass of urea required will be 37 g.

Ans: Given that,

20% of aqueous KI solution has 20 gm of KI in 100 gm of solution.

Mass of water (solvent) = 100 – 20 = 80 gm.

The volume of solution is given as,

Volume = $\dfrac{mass}{density}=\dfrac{100}{1.202}=83.194ml$ 

Molar mass of KI = 166.0028 g/mol

Thus,

Number of moles of KI = $\dfrac{20}{166.0028}=0.120mol$ 

Molar mass of water = 18 g/mol

Thus,

Number of moles of water = $\dfrac{80}{18}=4.44mol$ 

The molality of KI if the density of 20% (mass/mass) aqueous KI is 1.202 $g/ml$;

Molality = $\dfrac{0.120}{0.08}=1.50m$

1. The molarity of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g/ml.

Ans: Molarity of KI if the density of 20% (mass/mass) aqueous KI is 1.202 $g/ml$.

Molarity = $\dfrac{0.120}{0.083194}=1.44M$

1. Mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g/ml.

Ans: Mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 $g/ml$.

Mole fraction = $\dfrac{0.120}{0.120+4.44}=0.0263$

Ans: Molality (solubility) of ${{H}_{2}}S$in water = 0.195 m which states that,

0.195 moles of ${{H}_{2}}S$in 1Kg of water (1000 g).

Now,

Molar mass of water = 18 g/mol.

Number of moles of water = $\dfrac{1000}{18}=55.55mol$ 

Mole fraction of ${{H}_{2}}S$ is 

\[{{X}_{{{H}_{2}}S}}=\dfrac{0.195}{0.195+55.55}=0.00349\] 

Pressure at STP is given as 0.987 bar. Thus, applying Henry’s law we get;

\[{{P}_{{{H}_{2}}S}}={{K}_{H}}\times {{X}_{{{H}_{2}}S}}\] 

\[{{K}_{H}}=\dfrac{{{P}_{{{H}_{2}}S}}}{{{X}_{{{H}_{2}}S}}}=\dfrac{0.987}{0.00349}=282.80bar\] 

Therefore,  Henry’s constant will be 282.80 bar.

Ans: Given that,

Henry’s law constant, ${{K}_{H}}=1.67\times {{10}^{8}}Pa$

Pressure, ${{P}_{C{{O}_{2}}}}=2.5atm=2.5\times 101325=253312.5Pa$ 

Now, by Henry’s law;

\[{{P}_{C{{O}_{2}}}}={{K}_{H}}\times {{X}_{C{{O}_{2}}}}\] 

\[{{X}_{C{{O}_{2}}}}=\dfrac{253312.5}{1.67\times {{10}^{8}}}=1.5168\times {{10}^{-3}}\] 

Now,

500 ml soda water is equivalent to 500 ml water which indirectly signifies 500 gm of the same.

Thus,

We know that, the molar mass of water = 18 g/mol

Number of moles of water = $\dfrac{500}{18}=27.78mol$ 

Number of moles of $C{{O}_{2}}$ is given as,

\[{{X}_{C{{O}_{2}}}}=\dfrac{{{n}_{C{{O}_{2}}}}}{{{n}_{C{{O}_{2}}}}+{{n}_{water}}}\] 

\[{{n}_{C{{O}_{2}}}}={{X}_{C{{O}_{2}}}}\left( {{n}_{C{{O}_{2}}}}+{{n}_{water}} \right)\]

\[{{n}_{C{{O}_{2}}}}=42.20\times {{10}^{-3}}mol\]

Mass of $C{{O}_{2}}$ is given as,

Molar mass of $C{{O}_{2}}$ = 44 g/mol.

Thus,

Mass = ${{n}_{C{{O}_{2}}}}\times 44=1.857gm$

Ans: Vapor pressures of the given pure liquids are;

${{P}_{A}}^{0}=450mm$ Hg, 

${{P}_{B}}^{0}=700mm$ Hg and 

${{P}_{Total}}=600mm$ Hg 

By Raoult’s law;

\[{{P}_{Total}}={{P}_{A}}+{{P}_{B}}\] 

\[{{P}_{Total}}={{X}_{A}}{{P}_{A}}^{0}+{{X}_{B}}{{P}_{B}}^{0}={{X}_{A}}{{P}_{A}}^{0}+\left( 1-{{X}_{A}} \right){{P}_{B}}^{0}\]

\[{{P}_{Total}}={{P}_{B}}^{0}+\left( {{P}_{A}}^{0}-{{P}_{B}}^{0} \right){{X}_{A}}\] 

\[600 = 700 + (450-700){{X}_{A}}\]

\[\therefore {{X}_{A}}=0.4\] 

Thus,

\[{{X}_{B}}=1-0.4=0.6\] 

${{P}_{A}}={{X}_{A}}{{P}_{A}}^{0}=0.4\times 450=180mm$ Hg

${{P}_{B}}={{X}_{B}}{{P}_{B}}^{0}=0.6\times 700=420mm$ Hg 

Thus,

Mole fraction of A in vapor phase = $\dfrac{180}{180+420}=0.3$ 

Mole fraction of B in vapor phase = 1 – 0.3 = 0.7

Ans: Given that,

Vapor pressure of pure water, ${{P}^{0}}$ = 23.8 mm Hg

Mass of urea, ${{W}_{2}}$ = 50 g

Molar mass of urea, ${{M}_{2}}$ = 60 g/mol

Mass of water, ${{W}_{1}}$ = 850 g

Molar mass of water, ${{M}_{1}}$ = 18 g/mol

By Raoult’s law;

\[\Delta P={{P}^{0}}-{{P}_{s}}={{X}_{2}}\times {{P}^{0}}\] 

\[\dfrac{{{P}^{0}}-{{P}_{s}}}{{{P}^{0}}}={{X}_{2}}=\dfrac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}=\dfrac{{}^{{{W}_{2}}}/{}_{{{M}_{2}}}}{{}^{{{W}_{1}}}/{}_{{{M}_{1}}}+{}^{{{W}_{2}}}/{}_{{{M}_{2}}}}\]

\[{{X}_{2}}=0.017\] 

Now, 

We have given that ${{P}^{0}}$ = 23.8 mm Hg;

$\dfrac{23.8-{{P}_{s}}}{23.8}=0.017$ 

Thus, V. P. of solution, ${{P}_{s}}$  = 23.4 mmHg.

Ans: Given that,

Boiling point of water = 100${}^\circ $ C

Boiling point of water at 750 mm Hg = 99.63${}^\circ $ C

Elevation in boiling point, $\Delta {{T}_{b}}=100-99.63=0.37$ 

Ebullioscopic constant, ${{K}_{b}}=0.52$ 

Mass of water, ${{W}_{1}}$ = 500 g

Molar mass of water, ${{M}_{1}}$ = 18 g/mol

Molar mass of sucrose, ${{M}_{2}}$ = 342 g/mol

Mass of sucrose, ${{W}_{2}}=?$ 

The elevation in boiling pint is given as;

\[\Delta {{T}_{b}}={{K}_{b}}\times m\] 

\[\Delta {{T}_{b}}={{K}_{b}}\times \dfrac{{{n}_{2}}}{{{W}_{1}}}={{K}_{b}}\times \dfrac{{{W}_{2}}}{{{M}_{2}}}\times \dfrac{1000}{{{W}_{1}}}\] 

\[{{W}_{2}}=\dfrac{\Delta {{T}_{b}}\times {{M}_{2}}\times {{W}_{1}}}{{{W}_{2}}\times 1000}=121.67g\] 

$\therefore $ 121.67 g sucrose to be added in water such that it boils at 100${}^\circ $ C.