Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

Insulin contains $$3.4$$% sulphur. The minimum molecular mass of insulin is :

A
$$940$$

B
$$350$$

C
$$470$$

D
$$560$$

Solution

Correct answer: Option $$A$$

Explanation:
Since an atom is the smallest discrete unit of nature, a molecule of insulin must have at least one $$S$$ atom, it cannot have half an $$S$$ atom. Therefore, the minimum molecular mass of insulin will be when it has exactly one $$S$$ atom, which weights $$32g$$

$$\implies 3.4\% = 32g$$ of $$S$$
$$\implies 100\% = \dfrac{32}{3.4} \times 100\%$$
$$\implies 100\% = 941.17g$$
So Minimum molar mass of insulin is around $$940 g/mol$$

Hence, option $$A, 940$$ is correct.
Question 2
1 / -0

The molality of 1 L solution with x% $$H_2SO_4$$ is equal to 9. The weight of the solvent present in the solution is 910 g. The value of x in g per 100 mL is :

A
90

B
80.3

C
40.13

D
9

Solution

$$Molality=\dfrac {100\times \text {Weight of solute}}{Mw\times \text {Weight of solvent}}$$

$$9=\dfrac {1000\times W}{98\times 910}$$

$$W=802.6\ g\ L^{-1}$$ $$=80.26\ g\ per\ 100\ mL$$
Question 3
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$$13.4$$ g of a sample of unstable hydrated salt, $$Na_2SO_4.xH_2O$$ was found to contain $$6.3$$ g of $$H_2O$$. The number of molecules of water of crystallisation is :

A
$$5$$

B
$$7$$

C
$$2$$

D
$$10$$

Solution

Weight of salt $$=13.4$$ g
Weight of $$H_2O=6.3$$ g
Weight of anhydrous salt $$=13.4-6.3=7.1$$ g
Moles of anhydrous salt $$=\dfrac {7.1}{842}=0.05$$
Moles of $$H_2O=\dfrac {6.3}{18}=0.35$$ mol
$$0.05$$ mol of anhydrous salt gives $$0.35$$ mol of $$H_2O$$.
$$1$$ mol of anhydrous salt will give $$\dfrac {0.35}{0.05}=7$$ mol of water.
Question 4
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An experiment was set up to determine the percentage of water absorbed by raisins. If the mass of dry raisins was $$40 g$$ and mass of wet raisins was $$45 g$$, the percent water absorbed would be:

A
$$\displaystyle\frac{45}{40}\times 100$$

B
$$\displaystyle\frac{40}{45}\times 100$$

C
$$\displaystyle\frac{45 - 40}{40}\times 100$$

D
$$\displaystyle\frac{45 - 40}{45}\times 100$$

Solution

$${W}_{1} =$$ initial weight of raisins, $$40 g$$
$${W}_{2} =$$ final weight of wet raisins, $$45 g$$
$$\displaystyle\frac{{W}_{2} - {W}_{1}}{{W}_{1}}\times 100$$.

$$\displaystyle\frac{{45} - {40}}{{45}}\times 100$$
Question 5
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"Suvarnabhasm", an ayurvedic drug, is found to contain $$400$$ ppm of colloidal gold. Mass % of gold (atomic mass of Au $$=197$$ ) will be :

A
$$0.040\: %$$

B
$$7.88\: %$$

C
$$0.0788\: %$$

D
$$4\times 10^{-4}$$%

Solution

Mass % of gold means amount of gold present in $$ 100$$ g.
As it is given that, in $$10^6$$ g drug, gold present is $$ 400$$ g, so in $$ 100$$ g drug, gold is $$ \dfrac{400\times 100}{10^6} = 0.040$$ % of the compound.
Question 6
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Glucose is a physiological sugar. If the mass percentage of $$C=x\%$$, mass percent of $$H=y\%$$ and mass percent of $$O=z\%$$ in glucose $$(C_6H_{12}O_6)$$, then the value of $$x,\ y$$ and $$z$$ are, respectively:

A
$$40,\ 6.67,\ 53.33$$

B
$$30,\ 6.67,\ 43.33$$

C
$$40,\ 7.67,\ 63.33$$

D
none of the above

Solution

Molecular weight of glucose $$=C_6H_{12}O_6=12\times 6+1\times 12+16\times 6$$ $$=180$$ g
$$180$$ g of glucose contains $$72$$ g of $$C$$.
Percnetage of $$C=\dfrac {72\times 100}{180}=40\%$$

Percentage of $$H=\dfrac {12\times 100}{180}=6.67\%$$

Percentage of $$O=\dfrac {96\times 100}{180}=53.33\%$$
Question 7
1 / -0

If $$100$$ mL of $$H_2SO_4$$ and $$100$$ mL of $$H_2O$$ are mixed, the mass percent of $$H_2SO_4$$ in the resulting solution will be:
$$(d_{H_2SO_4}=0.09\: g\: mL^{-1}, d_{H_2O}=1.0\: g\: mL^{-1})$$

A
$$90.32$$%

B
$$47.36$$%

C
$$50.56$$%

D
$$60$$%

Solution

Mass of $${H_2SO_4}=\dfrac {100\times 0.9}{100\times 0.9+100\times 1}=\dfrac {90}{190}=0.4736$$ g.
% Mass of $$H_2SO_4=43.36$$%
Question 8
1 / -0

Atomic mass of an element is:

A
actual mass of one atom of the element

B
average mass of an atom of different atoms of the element

C
always a whole number

D
None of these

Solution

Atomic mass is an average mass of different atoms of an element, as most elements have different isotopes. Atomic mass is usually not a whole number. It can be a fraction.
Question 9
1 / -0

Law of conservation of mass was put forward by :

A
Lavoisier

B
Dalton

C
Priestly

D
Thomson

Solution

The Law of conservation of mass was put forward by Russian scientist Lomonosov in the year 1765. Later in 1783, French scientist Antoine Lavoisier also stated the same law independently.
Question 10
1 / -0

Under the same conditions, two gases have the same number of molecules. They must:

A
be noble gases

B
have equal volumes

C
have a volume of 22.4 dm$$^3$$ each

D
have an equal number of atoms

Solution

Avogadro's hypothesis: Equal volume of all gases have equal number of molecules (not atoms) at same temperature and pressures conditions.

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Some Basic Concepts of Chemistry Test - 26

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Some Basic Concepts of Chemistry Test - 26

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Question 1

$$940$$

$$350$$

$$470$$

$$560$$

Solution

Question 2

90

80.3

40.13

9

Solution

Question 3

$$5$$

$$7$$

$$2$$

$$10$$

Solution

Question 4

$$\displaystyle\frac{45}{40}\times 100$$

$$\displaystyle\frac{40}{45}\times 100$$

$$\displaystyle\frac{45 - 40}{40}\times 100$$

$$\displaystyle\frac{45 - 40}{45}\times 100$$

Solution

Question 5

$$0.040\: %$$

$$7.88\: %$$

$$0.0788\: %$$

$$4\times 10^{-4}$$%

Solution

Question 6

$$40,\ 6.67,\ 53.33$$

$$30,\ 6.67,\ 43.33$$

$$40,\ 7.67,\ 63.33$$

none of the above

Solution

Question 7

$$90.32$$%

$$47.36$$%

$$50.56$$%

$$60$$%

Solution

Question 8

actual mass of one atom of the element

average mass of an atom of different atoms of the element

always a whole number

None of these

Solution

Question 9

Lavoisier

Dalton

Priestly

Thomson

Solution

Question 10

be noble gases

have equal volumes

have a volume of 22.4 dm$$^3$$ each

have an equal number of atoms

Solution

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