# Solutions

## Chemistry

### NCERT

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.