jee main chemistry important questions | Solutions jee main questions FREE

Q 1 :
If a substance ‘A’ dissolves in solution of a mixture of ‘B’ and ‘C’ with their respective number of moles as $n_{A}, n_{B}$ and $n_{C}$ . Mole fraction of C in the solution is [2024]

$\frac{n_{C}}{n_{A} \times n_{B} \times n_{C}}$
$\frac{n_{C}}{n_{A} + n_{B} + n_{C}}$
$\frac{n_{C}}{n_{A} - n_{B} - n_{C}}$
$\frac{n_{B}}{n_{A} + n_{B}}$

1

2

3

4

(B)

Mole fraction of any substance = $\frac{Moles of that substance}{Total number of moles}$

Q 2 :
The Molarity (M) of an aqueous solution containing 5.85 g of NaCl in 500 mL water is:
(Given: Molar Mass Na : 23 and Cl : 35.5 g ${mol}^{- 1}$ ) [2024]

0.2
20
4
2

1

2

3

4

(A)

Assuming volume of solution same as volume of water,

$V_{solution} = 500 mL = 0.5 L$

$n_{solute} = \frac{W_{solute}}{M_{solute}} = \frac{5.85 g}{58.5 {g mol}^{- 1}} = 0.1 mol$

$M = \frac{n_{solute}}{V_{solution} (in L)} = \frac{0.1 mol}{0.5 L} = 0.2 M$

Q 3 :
The density of ‘x’ M solution (‘x’ molar) of NaOH is 1.12 g ${mL}^{- 1}$ , while in molality, the concentration of the solution is 3 m (3 molal). Then x is (Given: Molar mass of NaOH is 40 g/mol) [2024]

3.8
3.5
2.8
3.0

1

2

3

4

(D)

$molality = \frac{1000 \times Molarity}{1000 \times density - Molarity \times M_{solute}}$

$3 = \frac{1000 x}{1000 \times 1.12 - x \times 40}$

$3360 - 120 x = 1000 x$

$1120 x = 3360$

$x = 3$

Q 4 :
Molality (m) of 3 M aqueous solution of NaCl is:
(Given: Density of solution = 1.25 g ${mL}^{- 1}$ , Molar mass in g ${mol}^{- 1}$ : Na - 23, Cl - 35.5) [2024]

2.90 m
2.79 m
1.90 m
3.85 m

1

2

3

4

(B)

3M aqueous NaCl means:

3 mol NaCl in 1000 mL solution

$3 \times 58.5 g = 175.5 g NaCl in 1.25 {g mL}^{- 1} \times 1000 mL - 1250 g solution$

$3 mol NaCl in (1250 - 175.5) g = 1074.5 g = 1.074 kg solution$

$m = \frac{n_{solute}}{V_{solvent} (in kg)} = \frac{3}{1.0745} m = 2.79 m$

Q 5 :
Volume of 3 M NaOH (formula weight 40 g ${mol}^{- 1}$ ) which can be prepared from 84 g of NaOH is ______ $\times 10^{- 1}$ ${dm}^{3}$ . [2024]

0%

(7)

$Moles of NaOH (n_{NaOH}) = \frac{Given mass}{Molar mass} = \frac{84 g}{40 {g mol}^{- 1}} = 2.1 mol$

$Molarity of NaOH \times Volume of NaOH in L = Moles of NaOH$

$3 {mol L}^{- 1} \times Volume of NaOH in L = 2.1 mol$

$Volume of NaOH in L = 0.7 L or 0.7 {dm}^{3} = 7 \times 10^{- 1} {dm}^{3}$

Q 6 :
A solution of $H_{2} {SO}_{4}$ is 31.4% $H_{2} {SO}_{4}$ by mass and has a density of 1.25 g/mL. The molarity of the $H_{2} {SO}_{4}$ solution is _____ M (nearest integer)

[Given molar mass of $H_{2} {SO}_{4}$ = 98 g ${mol}^{- 1}$ ] [2024]

0%

(4)

$31.4 % H_{2} {SO}_{4} by mass means:$

$31.4 {gm H}_{2} {SO}_{4} in 100gm solution$

$\Rightarrow \frac{31.4}{98} mol = 0.32 {mol H}_{2} {SO}_{4} in \frac{100 g}{1.25 {g mL}^{- 1}}$

$= 80 mL solution$

$Molarity = \frac{n_{solute}}{V_{solution} (in mL)} \times 1000 = \frac{0.32}{80} \times 1000 M = 4 M$

Q 7 :
Molality of 0.8 M $H_{2} {SO}_{4}$ solution (density 1.06 g ${cm}^{- 3}$ ) is ______ $\times 10^{- 3}$ m. [2024]

0%

(815)

$m = \frac{M \times 1000}{d \times 1000 - M \times M_{solute}} = \frac{0.8 \times 1000}{1.06 \times 1000 - 0.8 \times 98} m$

$= 815 \times 10^{- 3} m$

Q 8 :
The mass of sodium acetate ( ${CH}_{3} COONa$ ) required to prepare 250 mL of 0.35 M aqueous solution is _____ g.
(Molar mass of ${CH}_{3} COONa$ is 82.02 g ${mol}^{- 1}$ ) [2024]

0%

(7)

$Volume of solution (V) = 250 mL = 0.25 L .$

${Moles of CH}_{3} COONa (n)$

$= Molarity \times volume = 0.35 \frac{mol}{L} \times 0.25 L = 0.0875 mol$

${Mass of CH}_{3} COONa$

$= n \times molar mass = 0.0875 mol \times 82.02 \frac{g}{mol} = 7.177 g$

Q 9 :
The molarity of 1 L orthophosphoric acid ( $H_{3} {PO}_{4}$ ) having 70% purity by weight (specific gravity 1.54 g ${cm}^{- 3}$ ) is ______ M.
(Molar mass of $H_{3} {PO}_{4}$ = 98 g ${mol}^{- 1}$ ) [2024]

0%

(11)

$70% w/w H_{3} {PO}_{4} solution means:$

$70 g H_{3} {PO}_{4} in 100 g solution$

$\Rightarrow \frac{70}{98} mol H_{3} {PO}_{4} in \frac{100}{1.54} mL solution$

$M = \frac{n_{solute}}{V_{solution} (in mL)} \times 1000$

$M = \frac{70 / 98}{100 / 1.54} \times 1000 M = \frac{70 \times 1.54 \times 1000}{98 \times 100} M = 11 M$

Q 10 :
Molality of an aqueous solution of urea is 4.44 m. Mole fraction of urea in solution is $x \times 10^{- 3}$ . Value of $x$ is _____ (integer answer) [2024]

0%

(74)

$m = \frac{x_{solute} \times 1000}{x_{solvent} \times M_{solvent}}$

$4.44 = \frac{x_{urea} \times 1000}{x_{water} \times 18}$

$\frac{x_{urea}}{x_{water}} = \frac{79.92}{1000}$

$x_{urea} = \frac{79.92}{79.92 + 1000} = 74 \times 10^{- 3}$

Topic Question Set

Q 1 :
If a substance ‘A’ dissolves in solution of a mixture of ‘B’ and ‘C’ with their respective number of moles as $n_{A}, n_{B}$ and $n_{C}$ . Mole fraction of C in the solution is [2024]

Q 2 :
The Molarity (M) of an aqueous solution containing 5.85 g of NaCl in 500 mL water is:
(Given: Molar Mass Na : 23 and Cl : 35.5 g ${mol}^{- 1}$ ) [2024]

Q 3 :
The density of ‘x’ M solution (‘x’ molar) of NaOH is 1.12 g ${mL}^{- 1}$ , while in molality, the concentration of the solution is 3 m (3 molal). Then x is (Given: Molar mass of NaOH is 40 g/mol) [2024]

Q 4 :
Molality (m) of 3 M aqueous solution of NaCl is:
(Given: Density of solution = 1.25 g ${mL}^{- 1}$ , Molar mass in g ${mol}^{- 1}$ : Na - 23, Cl - 35.5) [2024]

Q 5 :
Volume of 3 M NaOH (formula weight 40 g ${mol}^{- 1}$ ) which can be prepared from 84 g of NaOH is ______ $\times 10^{- 1}$ ${dm}^{3}$ . [2024]

0%

Q 6 :
A solution of $H_{2} {SO}_{4}$ is 31.4% $H_{2} {SO}_{4}$ by mass and has a density of 1.25 g/mL. The molarity of the $H_{2} {SO}_{4}$ solution is _____ M (nearest integer)

[Given molar mass of $H_{2} {SO}_{4}$ = 98 g ${mol}^{- 1}$ ] [2024]

0%

Q 7 :
Molality of 0.8 M $H_{2} {SO}_{4}$ solution (density 1.06 g ${cm}^{- 3}$ ) is ______ $\times 10^{- 3}$ m. [2024]

0%

Q 8 :
The mass of sodium acetate ( ${CH}_{3} COONa$ ) required to prepare 250 mL of 0.35 M aqueous solution is _____ g.
(Molar mass of ${CH}_{3} COONa$ is 82.02 g ${mol}^{- 1}$ ) [2024]

0%

Q 9 :
The molarity of 1 L orthophosphoric acid ( $H_{3} {PO}_{4}$ ) having 70% purity by weight (specific gravity 1.54 g ${cm}^{- 3}$ ) is ______ M.
(Molar mass of $H_{3} {PO}_{4}$ = 98 g ${mol}^{- 1}$ ) [2024]

0%

Q 10 :
Molality of an aqueous solution of urea is 4.44 m. Mole fraction of urea in solution is $x \times 10^{- 3}$ . Value of $x$ is _____ (integer answer) [2024]

0%

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Topic Question Set

Q 1 : If a substance ‘A’ dissolves in solution of a mixture of ‘B’ and ‘C’ with their respective number of moles as nA,nB and nC. Mole fraction of C in the solution is [2024]

Q 2 : The Molarity (M) of an aqueous solution containing 5.85 g of NaCl in 500 mL water is: (Given: Molar Mass Na : 23 and Cl : 35.5 gmol-1) [2024]

Q 3 : The density of ‘x’ M solution (‘x’ molar) of NaOH is 1.12 g mL-1, while in molality, the concentration of the solution is 3 m (3 molal). Then x is (Given: Molar mass of NaOH is 40 g/mol) [2024]

Q 4 : Molality (m) of 3 M aqueous solution of NaCl is: (Given: Density of solution = 1.25 g mL-1, Molar mass in g mol-1: Na - 23, Cl - 35.5) [2024]

Q 5 : Volume of 3 M NaOH (formula weight 40 g mol-1) which can be prepared from 84 g of NaOH is ______ ×10-1 dm3. [2024]

0%

Q 6 : A solution of H2SO4 is 31.4% H2SO4 by mass and has a density of 1.25 g/mL. The molarity of the H2SO4 solution is _____ M (nearest integer) [Given molar mass of H2SO4 = 98 g mol-1] [2024]

0%

Q 7 : Molality of 0.8 M H2SO4 solution (density 1.06 g cm-3) is ______ ×10-3 m. [2024]

0%

Q 8 : The mass of sodium acetate (CH3COONa) required to prepare 250 mL of 0.35 M aqueous solution is _____ g. (Molar mass of CH3COONa is 82.02 g mol-1) [2024]

0%

Q 9 : The molarity of 1 L orthophosphoric acid (H3PO4) having 70% purity by weight (specific gravity 1.54 g cm-3) is ______ M. (Molar mass of H3PO4 = 98 g mol-1) [2024]

0%

Q 10 : Molality of an aqueous solution of urea is 4.44 m. Mole fraction of urea in solution is x×10-3. Value of x is _____ (integer answer) [2024]

0%

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Q 1 :
If a substance ‘A’ dissolves in solution of a mixture of ‘B’ and ‘C’ with their respective number of moles as $n_{A}, n_{B}$ and $n_{C}$ . Mole fraction of C in the solution is [2024]

Q 2 :
The Molarity (M) of an aqueous solution containing 5.85 g of NaCl in 500 mL water is:
(Given: Molar Mass Na : 23 and Cl : 35.5 g ${mol}^{- 1}$ ) [2024]

Q 3 :
The density of ‘x’ M solution (‘x’ molar) of NaOH is 1.12 g ${mL}^{- 1}$ , while in molality, the concentration of the solution is 3 m (3 molal). Then x is (Given: Molar mass of NaOH is 40 g/mol) [2024]

Q 4 :
Molality (m) of 3 M aqueous solution of NaCl is:
(Given: Density of solution = 1.25 g ${mL}^{- 1}$ , Molar mass in g ${mol}^{- 1}$ : Na - 23, Cl - 35.5) [2024]

Q 5 :
Volume of 3 M NaOH (formula weight 40 g ${mol}^{- 1}$ ) which can be prepared from 84 g of NaOH is ______ $\times 10^{- 1}$ ${dm}^{3}$ . [2024]

Q 6 :
A solution of $H_{2} {SO}_{4}$ is 31.4% $H_{2} {SO}_{4}$ by mass and has a density of 1.25 g/mL. The molarity of the $H_{2} {SO}_{4}$ solution is _____ M (nearest integer)

[Given molar mass of $H_{2} {SO}_{4}$ = 98 g ${mol}^{- 1}$ ] [2024]

Q 7 :
Molality of 0.8 M $H_{2} {SO}_{4}$ solution (density 1.06 g ${cm}^{- 3}$ ) is ______ $\times 10^{- 3}$ m. [2024]

Q 8 :
The mass of sodium acetate ( ${CH}_{3} COONa$ ) required to prepare 250 mL of 0.35 M aqueous solution is _____ g.
(Molar mass of ${CH}_{3} COONa$ is 82.02 g ${mol}^{- 1}$ ) [2024]

Q 9 :
The molarity of 1 L orthophosphoric acid ( $H_{3} {PO}_{4}$ ) having 70% purity by weight (specific gravity 1.54 g ${cm}^{- 3}$ ) is ______ M.
(Molar mass of $H_{3} {PO}_{4}$ = 98 g ${mol}^{- 1}$ ) [2024]

Q 10 :
Molality of an aqueous solution of urea is 4.44 m. Mole fraction of urea in solution is $x \times 10^{- 3}$ . Value of $x$ is _____ (integer answer) [2024]