Class 12 Chemistry

Solid State

In Text Questions Part 4

Question - 2.1 - Calculate the mass percentage of benzene (C6 H 6) and carbon tetrachloride (CCl 4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Answer: Given, Mass of solute (WB ) = 22g

Mass of solvent (W B) = 122 g

Mass of solution = 22g + 122 g = 144 g

Mass % of benzene = ?

Mass % of carbon tetrachloride = ?

Now, Mass percentage of benzene `=(text(Mass of benzene)xx100)/text(Total mass of solution)`

`=(22)/(144)xx100=15.27%`

And, Mass percentage of carbon tetrachloride `=(text(Mass of CC)l_4xx100)/text(Total mass of solution)`

`=(122)/(144)xx100=84.72%`

Thus, Mass percentage of benzene = 15.27%

And, Mass percentage of carbon tetrachloride = 84.72%


Question – 2.2 - Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Answer: Given, % of benzene by mass = 30%

Thus, % of solvent (carbon tetrachloride) = 70%

Mole fraction of benzene in the given solution = ?

This means, 30g of benzene (C6 H 6) is dissolved in 70 g of CCl 4

Now, Molecular mass of benzene (C6 H6 ) = (12 x 6) + (1 x 6) = 78 g mol – 1

Molecular mass of carbon tetrachloride (CCl4 ) = 12 + (35.5 x 4 ) = 154 g mol – 1

Therefore, Number of moles of C 6H 6

`=text(Mass of benzene)/(text(Molecular mass of)C_6H_6)`

`=(30)/(78)=0.39` mol

And number of moles of CCl4 `=(text(Mass of CC)l_4)/(text(Molecular mass of CC)l_4)`

`=(70)/(154)=0.45` mol

Thus, mole fraction of benzene in the given solution

XBenzene = nBenzene ÷ (nBenzene + nCarbon tetrachloride

`=(0.38)/(0/38+0.45)=(0.38)/(0.83)=0.457`

Thus, mole fraction of benzene in the given solution = 0.457


Question – 2.3 - Calculate the molarity of each of the following solutions:

(a) 30 g of Co(NO 3) 2. 6H2 O in 4.3 L of solution

Answer: Given, Mass of solute (W B) = 30 g

Molar mass of given solute Co(NO 3) 2.6H 2O `= 58.7 + 2[14 + (16 xx 3)] + 6 ( 2 + 16)`

`= 58.7 + (2 xx 62) + (6 xx 18)`

`= 58.7 + 124 + 128` g mol – 1

= 290.7 g mol – 1

Now, Number of moles of Co(NO 3)2 .6H 2O

`=(text(Mass of cobalt nitrate)\W_B)/(text(Molecular mass of cobalt nitrate)\M_B)`

`=(30g)/(290.7g\text(mol)^(-1)=0.103` mol

Now, we know that, molarity `=text(Number of moles of solute)/text(Volume of solution in liter)`

`=(0.103text(mol))/(4.3L)=0.239 = 0.24` M

Thus, molarity of given solute = 0.24 M

(b) 30 mL of 0.5 M H 2SO4 diluted to 500 mL.

Answer: Given, 30 mL of 0.5 M H2SO4 diluted to 500 mL.

Thus, M1 = 30 mL, V1 = 0.5 M, V2 = 500 L, M2 = ?

We know that `M_1V_1=M_2V_2`

Or, `30xx0.5=M_2xx500`

Or, `M_2=(30xx0.5)/(500)=0.3` M

Thus, required molarity = 0.3 M

Alternate method:

Number of moles present in 1000 ml of 0.5 M H 2SO4 solution = 0.5 mol

Therefore, number of moles present in 1 ml of solution = 1 / 1000 mL

Therefore, number of moles present in 30 mL of solution `(1xx30)/(1000)=0.030`

Now, we know that, Molarity `=text(Number of moles of solute)/text(Volume of solutions in L)`

`=(0.030)/(0.5L)=0.06` M


Question – 2.4 - Calculate the mass of urea (NH2 CONH 2) required in making 2.5 kg of 0.25 molal aqueous solution.

Answer: Given, Molality (m) = 0.25 m

Molar mass (M B) of urea (NH2 CONH 2) = 14 + 2 + 12 + 16 + 14 + 2 = 60 g mol – 1

Mass of solvent (WA ) = 2.5 kg = 2500 gm

Mass of solute (W B) = ?

Now, we know that,

Molality (m) `=(W_B\xx1000)/(M_B\xx\W_A\gm)`

Or, `0.25=(W_B\xx1000)/(60g\text(mol)^(-1)xx2500gm)`

Or, `W_B=(0.25xx60xx2500)/(1000)`

Or, `W_B=37.50` g

Thus, mass of given solute urea = 37.50 g



Introduction

InText Questions

InText Que 2

InText Que 3