Mole Concept(11th And 12th > Chemistry ) Questions and answers for exam Preparation

11th And 12th > Chemistry

MOLE CONCEPT MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

Calculate the volume of $C O_{2}$ evolved at NTP when 20g of $C a C O_{3}$ is heated strongly.

3.38 L
4.48
5.58 L
6.68 L

Discuss Question

Answer: Option B. -> 4.48
:
B

The question gives weight of reactants and asks the volume of product. So, it is a weight-volume problem. Start by balancing the reaction

$C a C O_{3}$ $\to$ CaO + $C O_{2 (g)}$ $↑$

$∴$ 1 mole of $C a C O_{3}$ gives 1 mole of $C O_{2}$

From Avagadro's Law, 1 mole of $C O_{2}$ at NTP occupies 22.4L

$∴$ 1 mole of $C a C O_{3}$ = 22.4 L of $C O_{2}$
100g = 22.4L of $C O_{2}$

$∴$ 20g = 20 $\times$ $\frac{22.4}{100}$ = 4.48 L of $C O_{2}$

Question 2.

$\begin{matrix} I s o t o p e & R e l a t i v e A b u d a n c e (%) & A t o m i c M a s s (a m u)^{12} C & 98.892 & 12^{13} C & 1.108 & 13.00335^{14} C & 2 \times 10^{- 10} & 14.00317 \end{matrix}$

From the data given in the table, calculate the average atomic mass of C

13 u
12.011 u
12.001 u
12.0011 u

Discuss Question

Answer: Option B. -> 12.011 u
:
B

Average atomic mass is the weighted average of atomic mass of all the isotopes.
Contribution by each isotope = (Relative abundance) (Atomic mass)
Contribution by
$_{6}^{12} C$ = (0.98892)(12)/100 = 11.86704

$_{6}^{13} C$ = (0.01108)(13.00335)/100 = 0.14407712

$_{6}^{14} C$ = (2 $\times$ $10^{- 10}$ )(14.003170/100 = 28.00634 $\times$ $10^{- 12}$

$∴$ Average atomic mass = sum of all contribution = 12.011 u

Question 3.

An analytic balance has uncertainty In measurement equal to $\pm$ 1 mg. Then report the result in terms of percentage if the weight of a compound is
(a) 1 g
(b) 10 g
(c) 100 g

$1 \pm 0.1 %$ ; $10 \pm 0.01 %; 100$ $\pm 0.001 %$
$1 \pm 0.2 %; 10$ $\pm 0.02 %; 100$ $\pm 0.002 %$
$1 \pm 0.1 %; 10$ $\pm 0.1 %; 100$ $\pm 0.1 %$
$1 \pm 0.2 %; 10$ $\pm 0.2 %; 100$ $\pm 0.002 %$

Discuss Question

Answer: Option A. ->

1 \pm 0.1 %

;

10 \pm 0.01 %; 100

\pm 0.001 %

:
A

(a) $1 \pm 0.1 %$ (b) $10 \pm 0.01 %$ (c) $100 \pm 0.001 %$
The uncertainty in measurement is expressed in terms of percentage by putting $\pm$ sign before it, e.g., $150 \pm 1 %$ , et. Smaller the quantity to be measured, greater is the percentage uncertainty, and the instrument should be more precise for the measurement of smaller quantities.

$1 m g = \frac{1}{1000} g \frac{1}{1000} \times 100 = \frac{1}{10} = 0.1 1 \pm 0.1 %$

Question 4.

10g of Fe powder was mixed with a $C u S O_{4}$ solution which contained 70g of $C u S O_{4}$ . What is the weight of the copper obtained?

[Fe = 55.85g Cu = 63.6g]

10.39
11.39
12.39
13.39

Discuss Question

Answer: Option B. -> 11.39
:
B

The question talks about reaction between Fe and $C u S O_{4}$ . Observe that weights of the reactants are given and the weight of a product is asked. Thus, it is a weight-weight stoichiometry problem.
In all such stoichiometry problems, start by balancing the reaction
Fe + $C u S O_{4}$ $\to$ $F e S O_{4}$ + Cu
From the equation, 1 mole of Fe = 1 mole of Cu
$∴$ 55.85g $\approx$ 63.6g of Cu
∴ 10g of Fe will displace 11.39g of Cu if available.
The solution has 70g of $C u S O_{4}$
1 mole $C u S O_{4}$ has 63.6g of Cu
i.e., [63.6 + 32 + 16(4)]g of $C u S O_{4}$ = 63.6g of Cu
159.6g of $C u S O_{4}$ = 63.6g of Cu
70g of $C u S O_{4}$ = 27.76g of Cu

$∴$ 10 g = 10 $\times$ $\frac{63.6}{55.85}$ = 11.39 g of Cu

$∴$ In the $C u S O_{4}$ solution, 27.76g of Cu is present of which 11.39g of Cu will be precipitated out by the addition of 10g of Fe.

Question 5.

What is the n-factor / valency factor of $H_{2} C O_{3}$ ?

1
2
Either a or b
Neither a nor b

Discuss Question

Answer: Option C. -> Either a or b
:
C

In this case of $H_{2} C O_{3}$ valency factor changes depending upon the reaction as the number of $H^{+}$ ions replaced might vary depending on the extent of the reaction.

$H_{2} C O_{3}$ $\to$ $H^{+}$ + ${H C O_{3}}^{-}$ (n-factor =1)

$H_{2} C O_{3}$ $\to$ 2 $H^{+}$ + ${C O_{3}}^{2 -}$ (n-factor = 2)

Question 6.

What volume of 0.10 M $H_{2} S O_{4}$ must be added to 50 ml of a 0.10 M NaOH solution to make a solution in which the molarity of the $H_{2} S O_{4}$ is 0.050 M?

400 ml
50 ml
100 ml
150 ml

Discuss Question

Answer: Option B. -> 50 ml
:
B

2NaOH + $H_{2} S O_{4}$ $⟶$ $N a_{2} S O_{4}$ + 2 $H_{2} O$

0.0025 moles of $H_{2} S O_{4}$ will neutralise 0.005 moles NaOH.

So for 0.01 V moles H2SO4 there will be 0.01v-0.0025 moles H2SO4 left unused in the total volume(v+50ml) of solution .

The concentration here will be

$\frac{0.01 - 0.0025}{v + 50}$ $\times$ 1000 = 0.05M

$\frac{100 V - 2.5}{V + 50}$ =0.05

$\frac{100 V - 2.5}{V + 50}$ = 2.5+2.5

99.95V= 5

V= 0.05002L

V= 50.02ML

Question 7.

How to determine mole fraction of solvent when moles of solute and solvent are given?

$\frac{Moles of solute}{Moles of solvent}$
$\frac{Moles of solute}{Moles of solution}$
$\frac{Moles of solute}{Moles of solute+ Moles of solvent}$
$\frac{Moles of solvent}{Moles of solute+Moles of solvent}$

Discuss Question

Answer: Option D. ->

\frac{Moles of solvent}{Moles of solute+Moles of solvent}

:
D

Mole fraction (X) is just the ratio of the number of moles of the required quantity to the sum of number of moles of all other quantities. We usually only have a solute and a solvent.
$∴ X_{s o l v e n t} = \frac{M o l e s o f s o l v e n t}{Moles of solute + Moles of solvent}$