1   Calculate the mass percentage of benzene $(C_6 H_6 )$ and carbon tetrachloride $(CCl_4 )$ if $22 g$ of benzene is dissolved in $122 g$ of carbon tetrachloride.

Solution :

Mass percentage of $C_6 H_6 =\dfrac{\text{Mass of $C_6H_6$}}{\text{Total mass of the solution}}*100$%$\\$ $=\dfrac{\text{Mass of $C_6H_6$}}{\text{Mas of $C_6H_6$+ Mass of $CCl_4$}}*100$%$\\$ $=\dfrac{22}{22+122}*100$%$\\$ $=15.28$%$\\$ Mass percentage of $CCl_4$=$\\$ $\dfrac{\text{Mass of $CCl_4$}}{\text{Mass of $C_6H_6$+Mass of $CCl_4$}}*100$%$\\$ $=\dfrac{122}{22+122}*100$%$\\$ $=84.72$%$\\$ Alternatively,$\\$ Mass percentage of $CCl_4 = (100-15.28) = 84.72$%$\\$

2   Calculate the mole fraction of benzene in solution containing $30$% by mass in carbon tetrachloride.

Solution :

$30$% by mass of $C_6 H_6$ in $CCl_4 \Rightarrow 30 g C_6 H_6$ in $100 g$ solution$\\$ $\therefore $ Number of moles of $C_6 H_6 ( n^ C_6 H_6 ) = 30/78 = 0.385$ mol$\\$ (molar mass of $C_6 H_6 = 78g$)$\\$ No. of moles of $\\$ $CCl_4(n^CCl_4)=\dfrac{70}{154}=0.455\\ x_{C_{6}H_{6}}=\dfrac{n^C_6H_6}{n^C_6H_6+n^CCl_4}\\ =\dfrac{0.385}{0.385+0.455}\\ =\dfrac{0.385}{0.84}\\ =0.458\\ X_{CCl_{14}}=1-0.458=0.542$

3   Calculate the molarity of each of the following solution: $\\$ (a) $30 g$ of $Co(NO _3 ) _2 . 6H _2 O$ in $4.3 L$ of solution$\\$ (b) $30 mL$ of $ 0.5 M H_2 SO_4$ diluted to $500 mL.$

Solution :

(a) Molar mass of $Co(NO _3 )_ 2 .6H _2 O = 310.7 g mol ^{-1}$$\\$ no. of moles = $30/310.7 = 0.0966$$\\$ Vol. of solution $= 4.3 L$$\\$ Molarity $= 0.0966/4.3 = 0.022 M$$\\$ (b) $1000 mL $ of $0.5M H_2 SO_4$ contain $H_2 SO_4 = 0.5$ mole$\\$ $30 mL$ of $ 0.5 M H_2 SO_4$ contain $H_2 SO_4$$\\$ $= 0.5/1000 × 30 = 0.015$ mole $\\$ Volume of solution = $500mL = 0.5L$$\\$ Molarity =$ 0.015/0.5 = 0.03M$

4   Calculate the mass of urea $(NH_2 CONH_2 )$ required in making $2.5 kg$ of $0.25$ molal aqueous solution.

Solution :

$0.25$ molal aqueous solution to urea means that Moles of urea$ = 0.25$ mole$\\$ Mass of solvent $(NH_2 CONH_2 ) = 60 g$ mol $^{-1}$$\\$ $\therefore 0.25$ mole of urea $= 0.25 × 60 = 15g$$\\$ Mass of solution $= 1000 + 15g = 1.015 kg$$\\$ $1.015 kg$ of urea solution contains $15g$ of urea$\\$ $\therefore 2.5 kg$ of solution contains urea $= 15/1.015 × 2.5 = 37 g.$

5   Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of $20$% (mass/mass) aqueous $KI$ is $1.202 g mL ^{-1}$

Solution :

$20$% aq. $KOH$ solution $\Rightarrow 20g$ of $KI$ in $100g$ solution $\\$ $\therefore $ Mass of solvent $= 100 - 20= 80 g$$\\$ (i) Molality=$\dfrac{\text{no of moles of KI}}{\text{mass of solvent(kg)}}\\ =\dfrac{0.120}{0.080}=1.5 mol kg^{-1}$$\\$ (ii)Density of solution $= 1.202 g mL ^{-1}$$\\$ Volume of solution $=\dfrac{100}{1.202}=83.2 mL$$\\$ $=0.0832L$$\\$ $\therefore $ Molarity $= \dfrac{0.120}{0.0832}=1.44M$$\\$ (iii) No of moles of $Kl = 0.120$$\\$ $n_{H_2O}=\dfrac{80}{18}=4.44$$\\$ $x_{Kl}=\dfrac{0.120}{0.120+4.44}$$\\$ $=\dfrac{0.120}{4.560}=0.0263$