Some Basic Concepts of Chemistry Test

Result

Some Basic Concepts of Chemistry Test - 68

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Two beakers $$A$$ and $$B$$ present in a closed vessel. Beaker $$A$$ contains $$152.4\:g$$ aqueous solution of urea, containing $$12\:g$$ of urea. Beaker $$B$$ contains $$196.2\:g$$ glucose solution, containing $$18\:g$$ of glucose. Both the solutions allowed to attain the equilibrium. Determine weight $$\%$$ of glucose in its solution is at equilibrium.

A
$$\;25\%$$

B
$$\;10\%$$

C
$$\;18\%$$

D
$$\;15.5\%$$

Solution

The molar masses of glucose and urea are 180 g/mol and 60 g/mol respectively.
The number of moles of urea are $$\dfrac {12}{60} = 0.2$$
The number of moles of glucose are $$\dfrac {18}{180} = 0.1$$
When equilibrium is achieved, the mass ($$196.2 \ g$$) of glucose solution will reduce by x g and the mass of urea ($$152.4 \ g$$) will increase by x g. Due to this, the concentration (ratio of number of moles of solute to the mass of solution) of urea solution will become equal to the concentration of glucose solution.
$$\displaystyle \frac {0.2}{152.4+x} = \frac {0.1}{196.2-x} \\
152.4+x= 392.4-2x \\
3x = 240 \\
x = 80 $$
Total mass of glucose solution $$\displaystyle = 196.2-80 = 116.2$$ g
weight percent of glucose $$\displaystyle = \frac {18}{116.2} \times 100 =15.5$$ %
Question 2
1 / -0

The molality of $$1$$ L solution with $$x\%$$ $$H_2SO_4$$ is $$9$$. The weight of the solvent present in the solution is $$910$$ g. The value of $$x$$ is:

A
$$90$$

B
$$49$$

C
$$30$$

D
$$47$$

Solution

$$1$$ L of solution contains $$9$$ moles of sulfuric acid which corresponds to $$9 \times 98 = 882$$ g.
$$1$$ L of solution contains $$910$$ g water.
Hence, the percentage of sulphuric acid in the solution is $$\dfrac {882}{882+910}\times 100 = 49.2$$%.
Question 3
1 / -0

$$\displaystyle n_{1}$$ g of substance X reacts with $$\displaystyle n_{2}$$ g of substance Y to form $$\displaystyle m_{1}$$ g of substance R and $$\displaystyle m_{2}$$ g of substance of S. This reaction can be represented as $$X + Y \rightarrow R + S$$. The relation which can be established in the amounts of the reactants and the products will be :

A
$$\displaystyle n_{1}-n_{2}=m_{1}-m_{2}$$

B
$$\displaystyle n_{1}+n_{2}=m_{1}+m_{2}$$

C
$$\displaystyle n_{1}=n_{2}$$

D
$$\displaystyle m_{1}=m_{2}$$

Solution

The relation which can be established in the amounts of the reactants and the products will be as follows:

$$\displaystyle \text {Total mass of reactants}=\text {Total mass of products}$$

$$\displaystyle n_{1}+n_{2}=m_{1}+m_{2}$$

This is based on the law of conservation of mass.
Question 4
1 / -0

$$Na_2SO_4. \:xH_2O$$ has $$50\%$$ $$H_2O$$. Hence, $$x$$ is :

A
$$4$$

B
$$5$$

C
$$6$$

D
$$8$$

Solution

The molar mass of $$Na_2SO_4$$ is $$142$$ g/mol.
The molar mass of $$Na_2SO_4.xH_2O$$ is $$142+18x$$ g/mol. Since, $$50\%$$ water is present,
$$\displaystyle \dfrac {142+18x}{2}=18x$$
$$36x=142+18x$$
$$18x=142$$
$$x=7.9 \simeq 8$$
Question 5
1 / -0

Cortisone is a molecular substance containing $$21$$ atoms of carbon per molecule. The mass percentage of carbon in cortisone is $$69.98$$%. Its molar mass is:

A
$$176.5$$

B
$$252.2$$

C
$$287.6$$

D
$$360.1$$

Solution

Cortisone is a molecular substance containing $$21$$ atoms of carbon per molecule.
$$1$$ mole of cortisone contains $$21 \times 12 = 252$$ g of $$C$$.
The mass percentage of carbon in cortisone is $$69.98$$%.
Its molar mass will be $$ \dfrac {100}{69.98} \times 252 = 360.1$$ g/mol.
Question 6
1 / -0

The sodium salt of methyl orange has $$7\%$$ sodium. What is the minimum molecular weight of the compound?

A
$$420$$

B
$$375$$

C
$$329$$

D
$$295$$

Solution

Let $$M$$ be the minimum molecular weight of the compound.

$$7\%$$ of the compound is sodium. It is equal to $$\dfrac {7}{100} \times M = 0.07M$$ g sodium in $$1$$ mole of compound.

At least, $$1$$ mole of $$Na$$ will be present in $$1$$ mole of compound.
$$1$$ mole of sodium weighs $$23$$ g.

Thus, $$0.07M=23$$

$$M=\dfrac {23}{0.07}=329$$ g/mol

Thus, the minimum molecular weight of the compound is $$329$$ g/mol.

Hence, the correct option is $$C$$
Question 7
1 / -0

Mole fraction of ethanol and water mixture is $$0.25$$. Hence, percentage concentration of ethanol by weight of mixture is:

A
$$25\%$$

B
$$75\%$$

C
$$46\%$$

D
$$54\%$$

Solution

Explanation:
Molar mass of ethanol = $$46\ grams/(mole)$$
Molar mass of water = $$18\ gram/(mole)$$
The sum of all mole fractions is $$1$$.
The mole fraction of:
Ethanol = $$0.25$$ and water = $$0.75$$
Let the 10 moles of mixture are present.
Moles of ethanol $$=\dfrac {0.25}{10}\times 100$$ = $$2.5$$
Moles of water $$=\dfrac {0.75}{10}\times 100$$= $$7.5$$
Given mass of ethanol $$=2.5 \times 46 = 115\ grams$$
Given mass of water $$7.5 \times 18 = 135\ grams$$
The percentage concentration of ethanol by weight in the mixture $$=\dfrac {115}{115+135}\times 100$$
The percentage concentration of ethanol by weight in the mixture $$=46\%$$.
Hence, the correct answer is option $$C$$.
Question 8
1 / -0

A spherical ball of radius $$7$$ cm contains $$56\%$$ iron. If density is $$1.4$$ g/cc, the number of moles of $$Fe$$ present approximately is :

A
$$10$$

B
$$15$$

C
$$20$$

D
$$25$$

Solution

The volume of the ball is $$\dfrac {4}{3}\pi r^3 = \dfrac {4}{3} \times 3.1416 \times 7^3 = 1436.7$$ cc.
The density is $$1.4$$ g/cc.
The mass of the ball is $$1.4 \times 1436.7 = 2011.5$$ g.
However, the percentage of iron is $$56\%$$.
The mass of iron is $$\dfrac {56}{100}\times 2011.5 =1126.4$$ g.
The molar mass of $$Fe$$ is $$56$$ g/mol.
The number of moles of Fe present is $$\dfrac {1126.4}{56}=20$$.
Thus, the ball contains $$20$$ moles of $$Fe$$.
Question 9
1 / -0

Which of the following is a suitable example for illustrating the law of conservation of mass?
(Atomic mass of O = 16 g/mol, H = 1 g/mole)

A
$$18$$ g of water is formed by the combination of $$16$$ g oxygen with $$2$$ g of hydrogen.

B
$$18$$ g of water in liquid state is obtained by heating $$18$$ g of ice.

C
$$18$$ g of water is completely converted into vapour state on heating.

D
$$18$$ g of water freezes at $$4^oC$$ to give same mass of ice.

Solution

The option (A) is a suitable example for illustrating the law of conservation of mass.

18g of water is formed by the combination of 16g oxygen with 2g of hydrogen.

Total mass of reactants (O and H) = 16 + 2 = 18 g.
The mass of product (water) = 18 g.

Since, the mass of product is equal to the total mass of reactants, the mass is conserved during the reaction, thus illustrating the law of conservation of mass.

Note: The option (A) represents chemical change while all other options represent physical change. The law of conservation of mass corresponds to a chemical change.
Question 10
1 / -0

A salt is formed due to the reaction between an oxy acid containing chlorine and a base containing a monovalent metal of atomic mass $$x$$. The number of oxygen atoms in one molecule of the acid is more than the corresponding 'ic' acid. Calculate the molecular mass of the salt.

A
$$\displaystyle 83.5 + \frac{x}{4}$$

B
99.5 + $$x$$

C
83.5 + $$x$$

D
$$\displaystyle 99.5 + \frac{x}{2}$$

Solution

An oxy-acid containing chlorine with the name ending with 'ic' acid is chloric acid, $$HClO_3$$.

The number of oxygen atoms in one molecule of the acid is more than the corresponding 'ic' acid.

Hence, our oxyacid is perchloric acid $$HClO_4$$.

Let our base be $$MOH$$ where M is a monovalent metal of atomic mass $$x$$.

The acid base reaction is :

$$HClO_4 + MOH \rightarrow MClO_4 + H_2O$$

The atomic masses of metal, $$Cl$$ and $$O$$ are $$ x \ g/mol, 35.5 g/mol$$ and $$16 g/mol$$ respectively.

The molecular mass of the salt $$MClO_4$$ is $$x + 35.5 + 4 (16) = 99.5 + x$$

Hence, the correct answer is option $$\text{B}$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 68

Result

Some Basic Concepts of Chemistry Test - 68

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\;25\%$$

$$\;10\%$$

$$\;18\%$$

$$\;15.5\%$$

Solution

Question 2

$$90$$

$$49$$

$$30$$

$$47$$

Solution

Question 3

$$\displaystyle n_{1}-n_{2}=m_{1}-m_{2}$$

$$\displaystyle n_{1}+n_{2}=m_{1}+m_{2}$$

$$\displaystyle n_{1}=n_{2}$$

$$\displaystyle m_{1}=m_{2}$$

Solution

Question 4

$$4$$

$$5$$

$$6$$

$$8$$

Solution

Question 5

$$176.5$$

$$252.2$$

$$287.6$$

$$360.1$$

Solution

Question 6

$$420$$

$$375$$

$$329$$

$$295$$

Solution

Question 7

$$25\%$$

$$75\%$$

$$46\%$$

$$54\%$$

Solution

Question 8

$$10$$

$$15$$

$$20$$

$$25$$

Solution

Question 9

$$18$$ g of water is formed by the combination of $$16$$ g oxygen with $$2$$ g of hydrogen.

$$18$$ g of water in liquid state is obtained by heating $$18$$ g of ice.

$$18$$ g of water is completely converted into vapour state on heating.

$$18$$ g of water freezes at $$4^oC$$ to give same mass of ice.

Solution

Question 10

$$\displaystyle 83.5 + \frac{x}{4}$$

99.5 + $$x$$

83.5 + $$x$$

$$\displaystyle 99.5 + \frac{x}{2}$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.