Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 54

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Question 1
1 / -0

$$1g$$ of a sample of brass om reacting with excess HCI produces $$120 ml$$ of $${ H }_{ 2 }$$ gas at STP. The percentage Zn in this sample of brass is: (At.wt.of Zn-=$$65.5g$$)

A
$$32\%$$

B
$$35\%$$

C
$$38\%$$

D
$$40\%$$

Solution

$$Zn+2HCl\to ZnCl_2+H_2$$
Moles of $$H_2$$ released $$=\dfrac {120}{22400}=0.0053$$
$$\therefore \ $$ Moles of $$Zn=\dfrac {0.0053}{20}=5.3\ mole$$
$$\therefore \ \%$$ of $$Zn=\dfrac {5.3\times 10^{-3}\times 65.5}{1}\times 100$$
$$=35.7\ \%$$
Question 2
1 / -0

In $$500$$ grams of water dissolved 2 moles of potassium sulphate the mass precentage of salt in soluton:

A
$$41\%$$

B
$$60\%$$

C
$$35\%$$

D
$$48\%$$

Solution

Given that ,
$$2$$ moles of potassium sulphate is dissolved in $$500$$ gm of water
we know that ,
Molar mass of potassium sulphate$$= 174g$$
$$\therefore $$ Mass of potassium sulphate dissolved $$=2\times 174= 348\ g$$

Mass percentage of salt$$= \dfrac{\text{Mass of solute }}{\text{Mass of solution}}\times 100 \text %= \dfrac{348}{500+348}\times 100 \text %=\dfrac{348}{848}\times 100 \text %=41.03\text %=41\text %$$
Question 3
1 / -0

Calculate the mass of iron which will be converted into its oxide $$(Fe_3O_4)$$ by the action of $$8g$$ of steam on it. [At mass of $$Fe = 56$$]
$$Fe + H_2O \rightarrow Fe_3O_4 + 4H_2$$

A
$$21$$g

B
$$12$$g

C
$$42$$g

D
$$84$$g

Solution

$$3Fe+4H_{ 2 }O\rightarrow Fe_{ 3 }O_{ 4 }+4H_{ 2 }$$
As per stiochiometry, Moles of $$H_2O$$ =Moles of fe
$$4\times \cfrac{8}{18} =\cfrac{m}{56}\times 3$$
$$m=21 g$$
Therefore, mass of iron converted to $$Fe_3O_4$$ is $$21g.$$
Question 4
1 / -0

Vapour density of the equilibrium mixture of $${ NO }_{ 2 }$$ and $${ N }_{ 2 }{ O }_{ 4 }$$ is found to be 40 for the equilibrium:

$${ N }_{ 2 }{ O }_{ 4 }\rightleftharpoons 2{ NO }_{ 2 }$$.

Calculate the percentage of $${ NO }_{ 2 }$$ in the mixture?

A
10%

B
5%

C
26.08%

D
None of these

Solution

$$N_2O_4\rightleftharpoons 2NO_2$$

$$t=0$$ $$1$$

$$t=t_{eq}\ 1-\alpha \quad 2\alpha$$

Initial vapour density of N$$_2$$O$$_4$$ $$=\cfrac{92}{2}=46$$

Vapour density at equilibrium=40

$$\cfrac{46}{40}=\cfrac{c+c\alpha}{c}$$

$$\cfrac{46}{40}=1+\alpha$$

$$\dfrac {46}{1+\alpha}=40$$

$$\Rightarrow \ 46=40+40\alpha$$

$$\Rightarrow \ \alpha-\dfrac {6}{40}=\dfrac {3}{20}$$

$$\therefore \ \% \ NO_2=\dfrac {2\alpha}{1+\alpha}\times 100$$

$$=\dfrac {10}{23}\times 100+\dfrac {6}{23}\times 100 \times 26.08\%$$

Hence, the correct option is $$\text{C}$$
Question 5
1 / -0

A solution of phosphoric acid was made by dissolving 10.0 g $$H_3PO_4$$
in 100.0 mL water. The resulting volume was 104 mL. What is the , mole fraction of the solution

A
$$1$$

B
$$2$$

C
$$0.01$$

D
$$3$$

E
$$1.5$$
Question 6
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At $$0^{\circ}C$$ the density of gaseous oxide at 2 bar is same as that of nitrogen at 5 bar and $$0^{\circ}C$$. The molar mass of the oxide is (assuming ideal behaviour):

A
$$70 g/mol$$

B
$$140 g/mol$$

C
$$28 g/mol$$

D
$$60 g/mol$$

Solution

We know Density $$\rho= \cfrac {M_\rho}{RT} \longrightarrow (1)$$
For the given data if $$M$$ is the nolar mass of the gaseous oxide we have,
$$\rho= \cfrac {M}{RT}.2$$ or $$2= \cfrac {\rho RT}{M}\longrightarrow (2)$$
Also for nitrogen $$5=\cfrac {\rho RT}{28}\longrightarrow (3)$$
$$(2)$$ and $$(3) \Rightarrow \cfrac {5}{2}= \cfrac {M}{28}$$
or $$M= \cfrac {5\times 28}{2}=70g/mol$$
Question 7
1 / -0

Four one litre flasks are separately filled with gases $$O_2, F_2, CH_4$$ and $$CO_2$$ under same conditions.

The ratio of the number of molecules in these gases are:

A
$$2 : 2 : 4 : 3$$

B
$$1 : 1 : 1 : 1$$

C
$$1 : 2 : 3 : 4$$

D
$$2 : 2 : 3 : 4$$

Solution

Avogadro's law: It states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

$$\\ n_{1} : n_{2} : n_{3} : n_{4} = \cfrac{PV}{RT} : \cfrac{PV}{RT} : \cfrac{PV}{RT} : \cfrac{PV}{RT} = 1 : 1 : 1 : 1 $$

So, the ratio of no. of molecules $$= 1: 1: 1: 1$$

The correct option is $$B.$$
Question 8
1 / -0

What is the mass of the precipitate formed when $$50$$ mL of $$16.9$$ % $$(w/v)$$ solution of $$AgN{ O }_{ 3 }$$ is mixed with $$50$$ mL of $$5.8$$% $$(w/v)$$ $$NaCl$$ solution?

$$(Ag = 107.8, N =14, O = 16, Na = 23, Cl= 35.5)$$

A
$$7$$

B
$$14$$

C
$$28$$

D
$$3.5$$

Solution

$$AgNO_{ 3 }+NaCl\rightarrow AgCl\downarrow +NaNO_{ 3 }$$

Moles of $$AgNO_{ 3 }=\cfrac{\cfrac{16.9\times 50}{100}}{170}=0.05$$moles

Moles of $$NaCl =\cfrac{\cfrac{5.8\times 50}{100}}{58.5} =0.05$$ moles

Therefore, the moles of precipitate $$(AgCl)= 0.05$$ moles

Mass of precipitate formed $$(AgCl) =0.05\times 143.5=7.165g$$
Question 9
1 / -0

According to Avogadro's law the volume of a gas will ____ as _____ if ____ are held constant.

A
increases, number of moles; P & T

B
decreases, number of moles; P & T

C
increases; T & P; number of moles

D
decreases; P & T; number of moles

Solution

According to Avogadro's law: Equal volume of all gases at same temperature and pressure will have same no. of molecules.
OR
For a given mass of ideal gas,
Volume$$\propto$$Number of moles of the gas (if temperature and pressure are constant)
So, volume of the gas will increase as the number of moles if pressure (P) and temperature (T) are held constant.
Question 10
1 / -0

A 1.50 g sample of an ore containing silver was dissolved,and all the $$ Ag^{+} $$ was converted to 0.125 g $$ Ag_{2}S $$. What was the percentage of silver in the ore?

A
14.23%

B
8.27%

C
10.8%

D
7.2%

Solution

Number of moles of $$Ag_{2}S=\cfrac{0.125 g}{\text{Molecular weight of } Ag_{2}S}=\cfrac{0.125}{247.8}=5 \times 10^{-4} mole$$
Mass of silver$$= 5 \times 10^{4} mole \times 2 \times 108=0.108 g$$
$$\therefore$$Percentage of silver in ore$$=\cfrac{0.108}{1.5} \times 100 = 7.2 \%$$

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 54

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Some Basic Concepts of Chemistry Test - 54

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Question 1

$$32\%$$

$$35\%$$

$$38\%$$

$$40\%$$

Solution

Question 2

$$41\%$$

$$60\%$$

$$35\%$$

$$48\%$$

Solution

Question 3

$$21$$g

$$12$$g

$$42$$g

$$84$$g

Solution

Question 4

10%

5%

26.08%

None of these

Solution

Question 5

$$1$$

$$2$$

$$0.01$$

$$3$$

$$1.5$$

Question 6

$$70 g/mol$$

$$140 g/mol$$

$$28 g/mol$$

$$60 g/mol$$

Solution

Question 7

$$2 : 2 : 4 : 3$$

$$1 : 1 : 1 : 1$$

$$1 : 2 : 3 : 4$$

$$2 : 2 : 3 : 4$$

Solution

Question 8

$$7$$

$$14$$

$$28$$

$$3.5$$

Solution

Question 9

increases, number of moles; P & T

decreases, number of moles; P & T

increases; T & P; number of moles

decreases; P & T; number of moles

Solution

Question 10

14.23%

8.27%

10.8%

7.2%

Solution

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