Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

In what ratio should a 15% solution of acetic acid be mixed with a 3% solution of the acid to prepare a 10% solution?

[All percentages are mass/mass percentages.]

A
7:3

B
5:7

C
7:5

D
7:10
Question 2
1 / -0

$$Cu_{2}S$$ and $$M_{2}S$$ are isomorphous in which percentage of sulphur is $$20.14$$% and $$12.94$$% respectively. The atomic weight of M is:

[Atomic wt. of Cu = 63.5]

A
208

B
108

C
112

D
106

Solution

Assume atomic weight of $$S$$ and $$M$$ to be $$x$$ and $$y$$ respectively.

Percentage of sulfur is $$20.14$$% and $$12.94$$% in $$Cu_2S$$ and $$M_2S$$ respectively.

For $$Cu_2S$$, 2 moles Cu = 1 mole S.

1g of $$Cu_2S$$ contains 0.7986g Cu and 0.2014g S respectively.

Hence, $$2\times \dfrac {0.7986}{63.5} =1 \times \dfrac {0.2014 }{x}$$

1g of $$M_2S$$ contains 0.8706g M and 0.1294g S respectively.

Hence, $$2 \times \dfrac {0.8706}{y} =1 \times \dfrac {0.1294}{x} $$

Solving tow equation we get, $$y=107.70 \approx 108.$$
Question 3
1 / -0

A 1.24 M aqueous solution of KI has density of $$1.15 g / cm^{3}$$ . What is the percentage composition of solute in the solution?

A
$$17.89$$

B
$$27.89$$

C
$$37.89$$

D
$$47.89$$

Solution

1.24 M aqueous solution of KI means 1.24 moles of KI in 1 litre of solution.

Given,
Density of solution $$= 1.15\ \dfrac{g}{cm^3}$$

Volume of solution $$= 1\ L\ or\ 1000\ cm^3$$

$$\therefore$$ Mass of solution $$= Density\times volume$$

$$= 1.15\times 1000$$

$$= 1150\ g$$

Mass of solute $$= 1.24\times 166 = 205.8\ g$$

Percentage composition of solute $$= \dfrac{\text{Mass of solute}}{\text{ Mass of solution}}\times 100$$

$$= \dfrac{205.8}{1150}\times 100$$

$$= 17.89\%$$

Hence, option A is correct.
Question 4
1 / -0

When a certain quantity of oxygen was ozonised in suitable apparatus, the volume decreased by $$4\ ml$$. On addition of turpentine the volume further decreased by $$8\ ml$$. All volumes were measured at the same temperature and pressure. From these data, establish the formula of ozone.

A
$$\displaystyle \:O_{3}$$

B
$$\displaystyle \:O_{4}$$

C
$$\displaystyle \:O_{5}$$

D
$$\displaystyle \:O_{6.5}$$

Solution

Let the formula of ozone be $$\displaystyle O_n $$
When a certain quantity of oxygen was ozonised in suitable apparatus, the volume decreased by 4 ml.
$$\displaystyle nO_2 \rightleftharpoons 2O_n $$
$$\displaystyle n-2 \propto 4 $$
On addition of turpentine the volume further decreased by 8ml. All volumes were measured at the same temperature and pressure.
$$\displaystyle 8mL = 2\times 4mL $$
$$\displaystyle 2=2n-4 $$
$$\displaystyle 6=2n $$
$$\displaystyle n=3$$
Thus, the formula of ozone is $$\displaystyle O_3 $$.
Question 5
1 / -0

$$15\; g$$ of methyl alcohol is present in $$100$$ ml of solution. If density of solution is $$0.90$$ g/ml. Calculate the mass percentage of methyl alcohol in solution.

A
$$16.67\:\%$$

B
$$16.97\:\%$$

C
$$17.67\:\%$$

D
$$17.97\:\%$$

Solution

Mass of the solution present in $$100$$ ml of solution corresponds to $$100 \times 0.90 =90$$ g.

The mass percentage of the solution is $$\dfrac {\text {mass of methyl alcohol}}{\text {mass of solution}} \times 100 = \dfrac {15}{90} \times 100 = 16.67\%$$.

Hence, option A is correct.
Question 6
1 / -0

10 ml mixture of $$CO, CH_{4}$$ and $$N_{2}$$, exploded with an excess of oxygen, gave a contraction of 6.5 ml. There was a further contraction of 7ml when the residual gas was treated with $$KOH$$. Volume of $$CO, CH_{4}$$ and $$N_{2}$$ respectively, is :

A
5ml, 2ml, 3ml

B
2ml, 6ml, 4ml

C
1ml, 8ml, 9ml

D
4ml, 3ml, 5ml

Solution

$$\textbf{Solution}:$$
Let the volumes of $$CO, CH_4$$, and $$N_2$$ in the mixture be x ml, y ml, and (10-x-y) ml respectively. We ignore the volume of water formed in the combustion as it is a liquid and its volume is small. Nitrogen combusts only at very very high temperatures. So it remains unreactive.

$$2CO+O_2\implies2 CO_2$$ $$CH_4+2O_2\implies CO_2+2 H_2 O$$
x ml x/2 ml x ml y ml 2 y ml y ml (liquid)

The volume of oxygen used: $$2y+\dfrac{x}{2} ml$$
Total volume of all gases before combustion $$= 10 + 2 y + \dfrac{x}{2} ml$$
After combustion the total volume is: 10 ml
as $$CO_2 : x + y$$ $$N_2: 10 - x- y$$

Reduction in volume: $$10 + 2y + \dfrac{x}{2} - 10 = 2 y + \dfrac{x}{2} = 6.5 ml$$
$$\implies 4 y + x = 13 ml$$ --- (1)

When the mixture of $$CO_2 + N_2$$ passes over KOH, all of $$CO_2$$ is absorbed. Only $$N_2$$ remains. Then the reduction of 7 ml corresponds to that of $$CO_2$$.
$$\implies x + y = 7 ml$$ --- (2)

Solving the two equations, x = 5 ml and y = 2 ml

Volumes of $$C_O = 5 ml$$, $$CH_4 = 2 ml$$ , $$N_2 = 3 ml$$.

$$\textbf{Hence A is the correct option}$$
Question 7
1 / -0

20 ml of a mixture of $$C_{2}H_{2}$$ and CO was exploded with 30ml of oxygen. The gases after the reaction had a volume of 34ml. On treatement with $$KOH$$, 8ml of oxygen remained. Calculate the composition of the mixture.

A
$$C_{2}H_{2}= 6 ml, CO= 14 ml$$

B
$$C_{2}H_{2}= 4 ml, CO= 28 ml$$

C
$$C_{2}H_{2}= 8 ml, CO= 30 ml$$

D
$$C_{2}H_{2}= 10 ml, CO= 19 ml$$

Solution

Let the original composition of the mixture be $$x$$ ml $$C_2H_2$$ and $$20-x$$ ml $$CO$$.

The equations are as shown below:

$$\underset {x} {C_2H_2} +\underset {2.5x} {2.5O_2} \rightarrow \underset {2x} {2CO_2} +\underset {x} {H_2O} $$

$$\underset {20-x} {CO} +\underset {0.5(20-x)} {0.5O_2} \rightarrow \underset {20-x} {CO_2} $$

The volume of unreacted oxygen is 8 ml.

Hence, $$30 ml-2.5 x - 0.5(20-x) = 8 ml$$ or $$x=6 ml$$.

Hence, the composition of the mixture is 6 ml $$C_2H_2$$ and 14 ml $$CO$$.

Hence, the correct answer is option $$\text{A}$$.
Question 8
1 / -0

$$300$$ g of an aqueous solution of a particular solute (containing $$30$$% solute by mass) is mixed with $$400$$ g of another aqueous solution of the same solute (containing $$40$$% solute by mass). In the final solution, mass $$\%$$ of solute is :
[Given, Molecular mass of solute $$\displaystyle = 50$$]

A
$$40.09\%$$

B
$$39.87\%$$

C
$$34.05\%$$

D
$$35.71\%$$

Solution

$$300$$ g of $$30$$% solution contain $$300 \times \dfrac {30}{100}=90$$ g of solute.
$$400$$ g of $$40$$% solution contain $$400 \times \dfrac {40}{100}=160$$ g of solute.
The mass percent of the final solution $$= \dfrac {Mass\ of\ solute} {Mass\ of\ solution}=\dfrac {90+160} {300+400} \times 100=35.71\%$$.
Question 9
1 / -0

60ml of a mixture of nitrous oxide and nitric oxide was exploded with excess of hydrogen. If 38ml of $$N_{2}$$ was formed, calculate the volume of each gas in the mixture.

A
$$\displaystyle \:NO= 44ml; N_{2}O= 16ml$$

B
$$\displaystyle \:NO= 45ml; N_{2}O= 20ml$$

C
$$\displaystyle \:NO= 34ml; N_{2}O= 22ml$$

D
$$\displaystyle \:NO= 20ml; N_{2}O= 26ml$$

Solution

Let the given mixture contains $$2x$$ ml of $$NO$$. The volume of $$N_2O$$ present is $$60-2x$$ ml.

The balanced chemical equations are as shown below.

$$\underset {2x}{2NO}+2H_2 \rightarrow \underset {x}{N_2}+2H_2O $$

$$\underset {60-2x}{N_2O}+H_2 \rightarrow \underset {60-2x} {N_2}+H_2O $$

The volume of nitrogen collected is 38 ml.

Hence, $$60-2x+x=38$$

$$x=22 ml$$

Hence, the volumes of NO and $$N_2O $$ present in the originl mixture are $$2x=2(22)=44 ml$$ and $$60-2(22)=16 ml$$ respectively.

So, the correct option is $$A$$
Question 10
1 / -0

$$4.9$$ g sample of $$ KClO_{3}$$ was heated under such conditions that a part of it decomposed according to the equation: $$2KClO_{3}\rightarrow 2KCl+3O_{2}$$ and remaining underwent change according to the equation: $$4KClO_{3}\rightarrow 3KClO_{4}+KCl$$. If the amount of $$O_{2}$$ evolved was $$672$$ mL at $$1$$ atm and $$273$$ K, the percentage by weight of $$KClO_{4}$$ in the residue is :

A
$$52.72 \%$$

B
$$46.64\%$$

C
$$54.86\%$$

D
$$42.35\%$$

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 65

Result

Some Basic Concepts of Chemistry Test - 65

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SHARING IS CARING

Question 1

7:3

5:7

7:5

7:10

Question 2

208

108

112

106

Solution

Question 3

$$17.89$$

$$27.89$$

$$37.89$$

$$47.89$$

Solution

Question 4

$$\displaystyle \:O_{3}$$

$$\displaystyle \:O_{4}$$

$$\displaystyle \:O_{5}$$

$$\displaystyle \:O_{6.5}$$

Solution

Question 5

$$16.67\:\%$$

$$16.97\:\%$$

$$17.67\:\%$$

$$17.97\:\%$$

Solution

Question 6

5ml, 2ml, 3ml

2ml, 6ml, 4ml

1ml, 8ml, 9ml

4ml, 3ml, 5ml

Solution

Question 7

$$C_{2}H_{2}= 6 ml, CO= 14 ml$$

$$C_{2}H_{2}= 4 ml, CO= 28 ml$$

$$C_{2}H_{2}= 8 ml, CO= 30 ml$$

$$C_{2}H_{2}= 10 ml, CO= 19 ml$$

Solution

Question 8

$$40.09\%$$

$$39.87\%$$

$$34.05\%$$

$$35.71\%$$

Solution

Question 9

$$\displaystyle \:NO= 44ml; N_{2}O= 16ml$$

$$\displaystyle \:NO= 45ml; N_{2}O= 20ml$$

$$\displaystyle \:NO= 34ml; N_{2}O= 22ml$$

$$\displaystyle \:NO= 20ml; N_{2}O= 26ml$$

Solution

Question 10

$$52.72 \%$$

$$46.64\%$$

$$54.86\%$$

$$42.35\%$$

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