Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

The percentage of $${P}_{2}{O}_{5}$$ in diammonium hydrogen phosphate $$[{(N{H}_{4})}_{2}HP{O}_{4}]$$ is :

A
$$23.48 \%$$

B
$$53.78 \%$$

C
$$46.96 \%$$

D
$$71.0 \%$$

Solution

The molar mass of diammonium hydrogen phosphate is 132 g/mol and that of $$P_2O_5$$ is 142 g/mol.
1 mole of diammonium hydrogen phosphate contains 0.5 mol of $$P_2O_5$$.
132 g of diammonium hydrogen phosphate will contain 71 g of $$P_2O_5$$.
100 g of diammonium hydrogen phosphate will contain $$\dfrac {71}{132} \times 100 = 53.78$$ %
Hence, the percentage of $${P}_{2}{O}_{5}$$ in diammonium hydrogen phosphate $$[{(N{H}_{4})}_{2}HP{O}_{4}]$$ is $$53.78$$%
Question 2
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A $$1000$$ g sample of $$NaOH$$ contains $$3$$ moles of $$O$$ atom, what is the purity of $$NaOH$$?

A
$$14\%$$

B
$$100\%$$

C
$$12\%$$

D
$$24\%$$

Solution

A $$1000$$ g sample of $$NaOH$$ contains $$3$$ moles of $$O$$ atoms which corresponds to $$3$$ moles of $$NaOH$$.

The molar mass of $$NaOH$$ is $$40$$ g/mol.

$$3$$ moles of $$NaOH$$ weigh $$3 \times 40 = 120 $$ g.

Therefore, the purity of $$NaOH$$ is $$ \dfrac {120}{1000} \times 100=12$$%

Hence, the correct option is $$C$$
Question 3
1 / -0

The $$\%$$ loss in weight after heating a pure sample of potassium chlorate (Molecular weight $$= 122.5$$ g/mol) will be:

A
$$12.25$$

B
$$24.5$$

C
$$39.2$$

D
$$19.6$$

Solution

The balanced chemical equation for the thermal decomposition of potassium chlorate is,

$$2KClO_3 \xrightarrow {\Delta} 2KCl+3O_2$$

2 moles of potassium chlorate corresponds to 3 moles of oxygen.

Thus, $$2 \times 122.5 = 245 \: g$$ of potassium chlorate corresponds to $$3 \times 32 = 96 \: g$$ of oxygen.

The $$\%$$ loss in weight after heating a pure sample of potassium chlorate (Mol. wt. $$= 122.5$$ g/mol) will be $$\dfrac {96}{245} \times 100=39.2$$%.
Question 4
1 / -0

A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of $$15.0\%$$ by weight. $$3.0$$ g of the mineral on heating lost $$1.10$$ g of $$CO_2$$. The percent by weight of other metal is :

A
$$65$$

B
$$25$$

C
$$75$$

D
$$35$$

Solution

$$MCO_3\rightarrow MO+CO_2\uparrow$$
$$M'CO_3\rightarrow M'O+CO_2\uparrow$$
Equivalent of $$CO_2 =$$ Equivalent of carbonates of metals
$$1$$ mol of $$CO_2 \equiv 1$$ mol of $$CO_3^{2-}$$
$$44$$ g of $$CO_2 \equiv $$$$60$$ g of $$CO_3^{2-}$$
$$1.10$$ g of $$CO_2 \equiv \dfrac {60}{44}\times 1.10=1.5$$ g of $$CO_3^{2-}$$
% of $$CO_3^{2-}=\dfrac {1.5\times 100}{3}=50$$%
% of one metal $$ = $$ $$15$$ %
% of another metal $$ = 100-50-15=35$$%
Question 5
1 / -0

The simplest formula of a compound containing $$50$$% of element $$X$$ (Atomic mass $$=$$ $$10$$) and $$50$$% of the element $$Y$$ (Atomic mass $$=$$ $$20$$) by weight is:

A
$$XY$$

B
$$X_2Y$$

C
$$XY_2$$

D
$$X_2Y_3$$

Solution

Explanation:

Let us assume that the mass of compound is $$10$$ grams.

$$X$$ and $$ Y$$ both are $$50\%$$ of the mass of compound.

Therefore, mass of $$X$$ = mass of $$Y$$ = $$5$$ grams.

The atomic mass of $$X$$ = $$10$$ grams.

The atomic mass of $$Y$$ = $$20$$ grams.

$$Number \space of\space moles = \dfrac {given \space mass }{atomic \space mass}$$

$$Number \space of\space moles\ of \space X= \dfrac {5 }{10 }$$=$$0.5$$

$$Number \space of\space moles\ of \space Y= \dfrac {5}{20}$$ =$$0.25$$

The formula of compound is $$X_{0.5}Y_{0.25}$$ i.e. in simple form it is $$X_2Y$$.

Hence, the correct answer is option $$B$$.
Question 6
1 / -0

If the percentage of water of crystallization in $$MgSO_4.xH_2O$$ is $$13\%$$, then what is the value of $$x$$?

A
$$1$$

B
$$4$$

C
$$5$$

D
$$7$$

Solution

The molar mass of $$MgSO_4$$ is $$120.4$$ g/mol.

The molar mass of $$MgSO_4.xH_2O$$ is $$120.4+18x$$ g/mol.

Thus, $$1$$ mole of $$MgSO_4.xH_2O$$ weighs $$120.4+18x$$ g and contains $$18x$$ g water.

But, the percentage of water of crystallization in $$MgSO_4.xH_2O$$ is $$13\%$$.

Hence, $$\dfrac {13 \times (120.4+18x)}{100}=18x$$

or, $$120.4+18x=138.4x$$

or, $$120.4x=120.4$$

$$\therefore x=1$$

Thus, the value of $$x$$ is $$1$$.

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

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

A
$$25\%$$

B
$$75\%$$

C
$$46\%$$

D
$$54\%$$

Solution

$$\chi_2=\dfrac {n_2}{n_1+n_2}=0.25$$

$$\chi_1=\dfrac {n_1}{n_1+n_2}=0.75$$

$$\dfrac {\chi_1}{\chi_2}=\dfrac {n_1}{n_2}=\dfrac {3}{1}\Rightarrow \dfrac {W_1\times Mw_2}{Mw_1\times W_2}=\dfrac {3}{1}$$

$$\dfrac {W_1\times 46}{18\times W_2}=3$$

$$\dfrac {W_1}{W_2}=1.17$$

$$\therefore \dfrac {W_2}{W_1+W_2}=0.46$$

Therefore, % concentration of ethanol by weight in mixture $$=46$$%.

Hence, the correct option is $$C$$
Question 8
1 / -0

A hydrate of $$Na_2SO_3$$ has $$50\%$$ water by mass. It is :

A
$$Na_2SO_3\cdot 4H_2O$$

B
$$Na_2SO_3\cdot 6H_2O$$

C
$$Na_2SO_3\cdot 7H_2O$$

D
$$Na_2SO_3\cdot 2H_2O$$

Solution

$$Na_2SO_3:H_2O\equiv 50:50$$ by mass (Molecular weight of $$Na_2SO_3=126)$$
Number of moles $$= \dfrac {50}{126}:\dfrac {50}{18}$$
$$\therefore Ratio=1:7$$
Hence, the compound is $$Na_2SO_4.7H_2O$$.
Question 9
1 / -0

An aqueous solution of $$NaOH$$ having density $$1.1$$ $$kg/{dm}^{3}$$ contains $$0.02$$ mole fraction of $$NaOH$$. The $$\%$$ by mass of $$NaOH$$ in the solution is :

A
$$4.34$$

B
$$2.17$$

C
$$5.28$$

D
$$8.34$$

Solution

The mole fraction of NaOH is 0.02 mole.
Let,
Mass of water in the solution = 1000g
1000 g of water corresponds to 55.55 moles of water.

Mole fraction of solute $$=\displaystyle \frac {n}{55.55+n} = 0.02= \dfrac{1}{50}$$

$$\Rightarrow \displaystyle 50n = 55.55+n $$

$$\Rightarrow \displaystyle 49n = 55.55 $$

$$\Rightarrow \displaystyle n=\frac {55.55}{49} = 1.13$$ mol

Mass of NaOH $$=\displaystyle 1.13 \times 40 = 45.3$$ g.

$$\therefore$$ % by mass of NaOH $$=\displaystyle \frac {\text {Mass of NaOH}}{\text {Mass of NaOH} + \text {Mass of water}} \times 100$$

$$= \dfrac {45.3}{45.3+1000} = 4.34$$ %.

Option A is correct.
Question 10
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A mixture of $$CuSO_4.5H_2O$$ and $$MgSO_4.7H_2O$$ was heated until all the water was driven off. If $$5.0$$ g of mixture gave $$3$$ g of anhydrous salt, what was the percentage by mass of $$CuSO_4.5H_2O$$ in the original mixture?

A
$$74.4$$

B
$$70$$

C
$$80$$

D
$$90$$

Solution

Let the mixture is X% by mass of $$\displaystyle CuSO_4.5H_2O$$ and 100 - X % by mass of $$\displaystyle MgSO_4.7H_2O$$. 5.0 g of mixture will contain 0.05X g $$\displaystyle CuSO_4.5H_2O$$ and 5.0 - 0.05X g $$\displaystyle MgSO_4.7H_2O$$

The molar masses of $$\displaystyle CuSO_4.5H_2O$$ and $$\displaystyle MgSO_4.7H_2O$$ are 249.7 g/mol and 246.5 g/mol respectively.

The number of moles of $$\displaystyle CuSO_4.5H_2O = \dfrac {0.05X}{249.7} = 2.00 \times 10^{-4}X $$ moles.

The number of moles of $$\displaystyle MgSO_4.7H_2O = \dfrac {5.0-0.05X}{246.5}$$

Total number of moles of water obtained $$\displaystyle = 5 \times 2.00 \times 10^{-4}X + 7 \times \dfrac {5.0-0.05X}{246.5}
= 1.00 \times 10^{-3}X + \dfrac {5.0-0.05X}{35.22} $$

Mass of water obtained $$\displaystyle = 5.0 - 3.0 = 2.0 $$ g

Moles of water obtained $$\displaystyle = \dfrac {2.0}{18} = 0.11111$$
Hence,
$$\displaystyle 1.00 \times 10^{-3}X + \dfrac {5.0-0.05X}{35.22} = 0.11111 $$

$$\displaystyle 3.522 \times 10^{-2}X + 5.0-0.05X =3.9133 $$

$$\displaystyle 0.01478X =1.0867$$

$$\displaystyle X = 74.4$$ %

Hence, the percentage by mass of $$\displaystyle CuSO_4.5H_2O$$ in the original mixture was 74.4 %

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Some Basic Concepts of Chemistry Test - 67

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Some Basic Concepts of Chemistry Test - 67

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Question 1

$$23.48 \%$$

$$53.78 \%$$

$$46.96 \%$$

$$71.0 \%$$

Solution

Question 2

$$14\%$$

$$100\%$$

$$12\%$$

$$24\%$$

Solution

Question 3

$$12.25$$

$$24.5$$

$$39.2$$

$$19.6$$

Solution

Question 4

$$65$$

$$25$$

$$75$$

$$35$$

Solution

Question 5

$$XY$$

$$X_2Y$$

$$XY_2$$

$$X_2Y_3$$

Solution

Question 6

$$1$$

$$4$$

$$5$$

$$7$$

Solution

Question 7

$$25\%$$

$$75\%$$

$$46\%$$

$$54\%$$

Solution

Question 8

$$Na_2SO_3\cdot 4H_2O$$

$$Na_2SO_3\cdot 6H_2O$$

$$Na_2SO_3\cdot 7H_2O$$

$$Na_2SO_3\cdot 2H_2O$$

Solution

Question 9

$$4.34$$

$$2.17$$

$$5.28$$

$$8.34$$

Solution

Question 10

$$74.4$$

$$70$$

$$80$$

$$90$$

Solution

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