# Solutions

## NCERT Solution

### Part 3

Question 21: Two elements A and B form compounds having formula AB_{2} and AB_{4}. When dissolved in 20 g of benzene (C_{6}H_{6}) 1 g of AB_{2} lowers the freezing point by 2.3 K whereas 1.0 g of AB_{4} lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol^{-1}. Calculate atomic masses of A and B.

**Answer:** This problem can be solved by using following equation:

`M_2=(1000xx\k_f\xx\w_2)/(w_1xx\ΔT_f)`

So, `M_(AB_2)=(1000xx5.1xx1)/(20xx2.3)=110.87` g mol^{-1}

`M_(AB_4)=(1000xx5.1xx1)/(20xx1.3)=196.15` g mol^{-1}`

Let us assume that atomic masses of A and B are p and q

Then molar mass of `AB_2=p+2q=110.87` ………….(1)

Similarly, molar mass of `AB_4=p+4q=196.15` ……………..(2)

Subtracting equation (1) from (2) we get

`2q=85.28`

Or, `q=42.64`

Substituting the value of q in equation (1) we get:

`p+2xx42.64=110.87`

Or, `p+85.28=110.87`

Or, `p=110.87-85.28=25.59`

Hence, atomic mass of A = 25.59 g mol^{-1} and that of B = 42.64 g mol^{-1}

Question 22: At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

**Answer:** Molar mass of glucose = 180 g mol^{-1}

We know, `π=CRT`

So, `4.98=(W_1)/(M_1)xx\Rxx300`

Or, `4.98=(36)/(180)xxR\xx300`

Or, `4.98=60R` …………(1)

In second case, `1.52=C\xx\R\xx300` ……….(2)

Dividing equation (2) by (1), we get:

`(1.52)/(4.98)=(C\xx\R\xx300)/(60R)`

Or, `5C=(1.52)/(4.98)`

Or, `C=(1.52)/(4.98xx5)=0.06` M

Question 23: Suggest the most important type of intermolecular attractive interaction in the following pairs:

- N-hexane and n-octane
- I
_{2}and CCl_{4} - NaClO
_{4}and water - Methanol and acetone
- Acetonitrile (CH
_{3}CN) and acetone (C_{3}O_{6}O)

**Answer:** (a) London dispersion forces, (b) London dispersion forces, (c) Ion-dipole interactions, (d) Dipole-dipole interactions

Question 24: Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH_{3}OH, CH_{3}CN.

**Answer:** Cylcohexane mixes in all proportionas, KCl does not dissolve in n-octane, methanol dissolves, `CH_3CN` dissolves to a greater extent than methanol

Question 25: Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (i) phenol, (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform (vi) pentanol

**Answer:** (i) Partially soluble, (ii) Insoluble, (iii) Highly Soluble, (iv) Highly soluble, (v) Insoluble, (vi) Partially soluble

Question 26: If the density of some lake water is 1.25 g mL^{-1} and contains 92 g of Na^{+} ions per kg of water, calculate the molality of Na^{+} ions in the lake.

**Answer:** Molar mass of na = 23 g mol^{-1}

So, number of moles of Na^{+} ions in 1 kg of water

`=(92)/(23)=4` mole

Hence, molality `=(4xx1000g)/(1000g) = 4` m

Question 27: If the solubility of CuS is 6 `xx` 10^{-16}, calculate the maximum molarity of CuS in aqueous solution.

**Answer:** Maximum molarity of CuS means solubility of Cus

Let us assume that solubility of CuS is S mol L^{-1}

Hence, `K_(sp)=[Cu^(2+)][S^(2-)]`

Or, `6xx10^(-16)=S\xx\S=S^2`

Or, `S=sqrt(6xx10^(-16))=2.45xx10^-8` mol L^{-1}

Question 28: Calculate the mass percentage of aspirin (C_{9}H_{8}O_{4}) in acetonitrile (CH_{3}CN) when 6.5 g of C_{9}H_{8}O_{4} is dissolved in 450 g of CH_{3}CN.

**Answer:** Mass percentage of aspirin

`=text(Mass of aspirin)/(text(Mass of aspirin)+text(Mass of acetonitrile))xx100`

`=(6.5)/(6.5+450)xx100=(6.5)/(456.5)=1.424%`

Question 29: Nalorphene (C_{19}H_{21}NO_{3}), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Does of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 `xx` 10^{-3} m in aqueous solution required for the above dose.

**Answer:** Molar mass of nalorphene

`=19xx12+21+14+3xx16=228+21+14+48=311` g mol^{-1}

So, mass of given mole of nalorphene `=1.5xx10^(-3)xx311=0.467` g

Hence, mass of given solution `=0.467+1000=1000.467` g

Since for 0.467 g of nalorphen, mass of required solution = 1000.467 g

Hence, for 1.5 mg `= 1.5 xx 10^(-3)` g of nalorphene, mass of required solution

`=(1000.467)/(0.467)xx1.5xx10^(-3)=3.21` g

Question 30: Calculate the amount of benzoic acid (C_{6}H_{5}COOH) required for preparing 250 mL of 0.15 M solution in methanol.

**Answer:** Molar mass of benzoic acid

`=6xx12+5xx1+12+2xx16+1=72+5+12+32+1=122` g mol^{-1}

1 L of 1 M benzoic acid solution will contain 122 g benzoic acid

So, 1 L of 0.15 M benzoic acid solution will contain `122xx0.15=18.3` g benzoic acid

Hence, mass of benzoic acid in 250 mL solution `=(18.3)xx(250)/(1000)=4.575` g of benzoic acid

Question 31: The depression in freezing point of water observed for the same amount of acetic acid, tricholoroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

**Answer:** Fluorine is more electronegative than chlorine. Hence, fluorine has higher electron withdrawing inductive effect than chlorine. Due to this, trifluoroacetic acid is the strongest acid, followed by trichloroacetic acid and then by acetic acid. Due to this, trifluoroacetic acid ionizes to the largest extent, while acetic acid ionizes to the least extent. So, depression in freezing point of water is least for acetic acid and greatest for trifluoroacetic acid.