MCQ Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 12

When 100 ml of a

$O_{2} - O_{3}$ mixture was passed through turpentine, there was the reduction of volume by 20 ml. If 100ml of such a mixture is heated, what will be the increase in volume?

0%
$10\ ml$
0%
$20\ ml$
0%
$30\ ml$
0%
$25\ ml$

Explanation

Ozone is highly soluble in turpentine. Thus, the decrease in volume of

$20$ ml is due to the solubility of ozone. Thus

$100 ml$ of a mixture contains

$20$ ml of ozone.

On heating,

$20$ ml of ozone decomposes to form

$30$ ml of oxygen. The net increase in volume will be

$(30 - 20)=10 ml$ .

$2O_3 \rightarrow 3O_2$

So, the correct option is

$A$

The average atomic mass of a mixture containing 79 mole percent of

$\displaystyle ^{24}\textrm{Mg}$ and remaining 21 mole percent of

$\displaystyle ^{25}\textrm{Mg}$ and

$\displaystyle ^{26}\textrm{Mg}$ , is 24.Mole percent of

$\displaystyle ^{26}\textrm{Mg}$ is :

0%
5
0%
20
0%
10
0%
15

Explanation

Average atomic mass

$= 24.31$

Mole percentage of

$\displaystyle ^{24}\textrm{Mg}$ is

$79$ %.

Let mole percentage of

$\displaystyle ^{26}\textrm{Mg}$ be

$x$ .

Mole percentage of

$\displaystyle ^{25}\textrm{Mg}$ will be

$21-x$ .

$24.31 = \dfrac {0.79 \times 24 + 0.01x \times 26 +0.01 \times (21-x) \times 25} {0.79+0.01x+0.01(21-x)}$ .
Hence,

$x=10$ %.

$15\; g$ 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.

0%
$16.67\:\%$
0%
$16.97\:\%$
0%
$17.67\:\%$
0%
$17.97\:\%$

Explanation

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.

Calculate percentage change in

$M_{avg}$ of the mixture, if

$PCl_{5}$ undergo

$50\%$ decomposition is a closed vessel.

0%
$50\%$
0%
$66.66\%$
0%
$33.65\%$
0%
$Zero$

Explanation

This is because dissociation of one molecule of

$PCl_5$ gives one molecule of

$PCl_3$ and one molecule of

$Cl_2$ .

Also,

$M_{PCl_5}=M_{PCl_3}+M_{Cl_2}$

Let, initially 2n moles of

$PCl_5$ be present.

Then,

$M_{avg}= \dfrac {2n \times 208.5} {2n} =208.5\ g/mol$

After

$50$ % dissociation, the mixture contains n moles of

$PCl_5$ , n moles of

$PCl_3$ and n moles of

$Cl_2$ .

Then

$M_{avg}= \dfrac {n \times 208.5 + n \times 137.5 + n \times 71} {3n} =138.33\ g/mol$

The percentage change in

$M_{avg}$ of the mixture

$= \dfrac {208.5-138.33} {208.5} \times 100 = 33.65$ %

$300$ 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

$\displaystyle = 50$ ]

0%
$40.09\%$
0%
$39.87\%$
0%
$34.05\%$
0%
$35.71\%$

Explanation

$300$ 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

$= \dfrac {Mass\ of\ solute} {Mass\ of\ solution}=\dfrac {90+160} {300+400} \times 100=35.71\%$ .

The mole fraction of solute in aqueous urea solution is

$0.2$ . The mass percent of solute is:

0%
$45.45\%$
0%
$43.34\%$
0%
$34.23\%$
0%
$35.67\%$

Explanation

The mole fraction of urea is

$0.2$ .

The mole fraction of water is

$1-0.2=0.8$ .

The molar masses of urea and water are

$60$ g/mol and

$18$ g/mol respectively.

Let us say one mole of the total solution is present.

The number of moles of urea and water present will be

$0.2$ mol and

$0.8$ mol. Their masses will be

$0.2 \times 60 = 12$ g urea and

$0.8 \times 18=14.4$ g water.

The mass percentage of solution

$= \dfrac {Mass\ of \ urea} {Total \ mass\ of\ solution} = \dfrac {12} {12+14.4} \times 100=45.45$ %.

Which of the following statements is correct about the reaction given below?

$\displaystyle 4Fe\left ( s \right )+3O_{2}\left ( g \right )\rightarrow 2Fe_{2}O_{3}\left ( g \right )$

0%
The total mass of reactants $=$ Total mass of the products. It follows the law of conservation of mass.
0%
The total mass of reactants $=$ Total mass of the products. Therefore, the law of multiple proportions is followed.
0%
Amount of $Fe_{2}O_{3}$ can be increased by taking any one of the reactants (iron or oxygen) in excess.
0%
Amount of $Fe_{2}O_{3}$ produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.

Explanation

Total mass of iron and oxygen in reactants

$=4 (55.85) + 3(32) =319.4\ g$ .

Total mass of iron and oxygen in product

$=2(2 \times 55.85 + 3 \times 8) =319.4\ g$ .

The total mass of iron and oxygen in reactants = the total mass of iron and oxygen in the product, therefore, it follows the law of conservation of mass.

Hence, option

$A$ is correct.

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 :

0%
5ml, 2ml, 3ml
0%
2ml, 6ml, 4ml
0%
1ml, 8ml, 9ml
0%
4ml, 3ml, 5ml

Explanation

$\textbf{Solution}:$

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

$CO_2 : x + y$

$N_2: 10 - x- y$

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

$C_O = 5 ml$ ,

$CH_4 = 2 ml$ ,

$N_2 = 3 ml$ .

$\textbf{Hence A is the correct option}$

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.

0%
$\displaystyle \:NO= 44ml; N_{2}O= 16ml$
0%
$\displaystyle \:NO= 45ml; N_{2}O= 20ml$
0%
$\displaystyle \:NO= 34ml; N_{2}O= 22ml$
0%
$\displaystyle \:NO= 20ml; N_{2}O= 26ml$

Explanation

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

$A$

How many moles of

$Na^{+}$ ions are present in

$20$ mL of

$0.40$ M

$Na_3PO_4$ ?

0%
$0.0080$
0%
$0.024$
0%
$0.050$
0%
$0.20$

Explanation

$1$ molecule of

$Na_3PO_4$ dissociates to give

$3$ sodium ions.
The number of moles of

$Na^{+}$ ions present in

$20$ mL of

$0.40\ M$

$Na_3PO_4$

$=\dfrac {20}{1000} \times 0.40 \times 3 = 0.024$ .
Thus,

$0.024$ mol of

$Na^{+}$ ions are present in

$20$ mL of

$0.40\ M$

$Na_3PO_4$ .

Two gases A and B which react according to the equation

$\displaystyle \:aA_{(g)}+bB_{(g)}\rightarrow cC_{(g)}+dD_{(g)}$ to give two gases C and D are taken (amount not known) in an Eudiometer tube (operating at a constant pressure and temperature) to cause the above. If on causing the reaction there is no volume change observed then which of the following statement are correct :

0%
$\displaystyle \:\left ( a+b \right)= \left ( c+d \right )$
0%
Average molecular mass may increase or decrease if either of A or B is present in limited amount.
0%
Vapour Density of the mixture will remain same throughout the course of reaction.
0%
Total moles of all the component of mixture will change.

10 ml of CO is mixed with 25 ml air having 20 per cent

$O_{2}$ by volume. What would be the final volume if none of CO and

$\displaystyle\: O_{2}$ is left after the reaction?

0%
30 ml
0%
40 ml
0%
43 ml
0%
49 ml

Explanation

The air contains approximately 20% oxygen and 80% nitrogen.

Thus 25 ml of air contains 5 ml oxygen and 20 ml nitrogen.

The reaction for the conversion of CO is as shown below.

$\underset {10 ml} {CO} + \underset {5 ml} {\cfrac {1} {2}} O_2 \rightarrow \underset {10 ml} {CO_2}$

Hence, the volume of the mixture after reaction is

$10\ ml\ CO_2 + 20\ ml\ N_2 =30\ ml.$

Hence, the correct answer is option

$\text{A}$ .

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.

0%
$\displaystyle \:O_{3}$
0%
$\displaystyle \:O_{4}$
0%
$\displaystyle \:O_{5}$
0%
$\displaystyle \:O_{6.5}$

Explanation

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

$\displaystyle O_3$ .

$40\: g\: Ba(MnO_{4} )_{2}$ (mol. wt.

$= 375$ g/mol) sample containing some inert impurities in acidic medium is completely reacted with

$125$ mL of

$33.6\: V$ of

$H_{2}O_{2}$ . What is the percentage purity of the sample?

0%
$28.12 \%$
0%
$70.31 \%$
0%
$85 \%$
0%
None of the above

Explanation

Mass of

$H_2O_2$ in

$125$ ml solution

$=\dfrac{17}{5600}\times 33.6 \times 125$ g
So, number of equivalents of

$H_2O_2 = 33.6 \times \dfrac{125}{5600}$
As number of equivalents of both are same,

$\dfrac{x}{\dfrac{375}{10}} = 33.6\times \dfrac{125}{5600}$

or,

$x= 28.125$
So, percentage purity of the sample

$= \dfrac{28.125}{40}\times 100 = 70.31\%$

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.

0%
$C_{2}H_{2}= 6 ml, CO= 14 ml$
0%
$C_{2}H_{2}= 4 ml, CO= 28 ml$
0%
$C_{2}H_{2}= 8 ml, CO= 30 ml$
0%
$C_{2}H_{2}= 10 ml, CO= 19 ml$

Explanation

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

$\text{A}$ .

Calculate the volume of

$CO_{2}$ evolved by the combustion of

$50$ ml of a mixture containing

$40$ per

$C_{2}H_{4}$ and

$60$ per

$CH_{4}$ (by volume).

0%
$70$ ml
0%
$75$ ml
0%
$80$ ml
0%
$82$ ml

Explanation

The volume of

$C_2H_4$ in

$50$ ml of sample is

$50 ml \times \dfrac {40} {100} =20 ml$ .

The volume of

$CH_4$ in

$50$ ml of sample is

$50 ml \times \dfrac {60} {100} =30 ml$ .

$\underset {30 ml}{CH_4} \rightarrow \underset {30 ml}{CO_2} + \underset {60 ml} {2H_2O}$

$\underset {20 ml}{C_2H_4} \rightarrow \underset {40}{2CO_2} + \underset {40 ml} {2H_2O}$
The volume of carbon dioxide formed is

$30 ml + 40 ml=70 ml$ .

A mixture of

$NH_{4}NO_{3}$ and

$(NH_{4})_{2}HPO_{4}$ contains

$30.40 \%$ mass percent of nitrogen. What is the mass ratio of the two components in the mixture?

0%
$2:1$
0%
$1:2$
0%
$3:4$
0%
$4:1$

Explanation

Let weight of

$NH_{4}NO_{3}$ and

$(NH_{4})_{2}HPO_{4}$ are

$x$ and

$y$ g respectively.
So, mass per cent of nitrogen is given by,

$\displaystyle \frac{\displaystyle \frac{x}{80} \times 2 \times 14+\displaystyle \frac{y}{132} \times 2 \times 14}{x+y} \times 100=30.4$

$\Rightarrow x:y=2:1$

10 ml of a mixture of

$CH_{4}, C_{2}H_{4}$ and

$CO_{2}$ were exploded with excess of air. After explosion, there was contraction on cooling of 17 ml and after treatment with KOH, there was further reduction of 14 ml. What is the composition of the mixture?

0%
$CH_{4}= 4.5 ml, CO_{2}= 1.5 ml$
0%
$CH_{4}= 3.6 ml, CO_{2}= 2.5 ml$
0%
$CH_{4}= 4.8 ml, CO_{2}= 1.8 ml$
0%
$CH_{4}= 5.1 ml, CO_{2}= 2.8 ml$

Explanation

The contraction on cooling is due to the condensation of water vapour. Thus 17 ml of water are formed in the reaction.

The contraction in the volume of 14 ml is due to the dissolution of carbon dioxide gas. Hence, 14 g of carbon dioxide is present in the mixture after the explosion.

$\underset {4.5 ml}{CH_4} \rightarrow \underset {4.5 ml}{CO_2} + \underset {9 ml} {2H_2O}$

$\underset {4 ml}{C_2H_4} \rightarrow \underset {8}{2CO_2} + \underset {8 ml} {2H_2O}$

Hence, the composition of the mixture is 4.5 ml methane and 1.5 ml of carbon dioxide.

40 ml of mixture of

$C_{2}H_{2}$ and CO is mixed with 100 ml of

$O_{2}$ gas and the mixture is exploded. The residual gases occupied 104 ml and when these are passed through KOH solution, the volume becomes 48ml. All the volume are at same the temperature and pressure. Determine the composition of the original mixture.

0%
$\displaystyle\:C_{2}H_{2}= 16 ml, CO= 24 ml$
0%
$\displaystyle\:C_{2}H_{2}= 23 ml, CO= 14 ml$
0%
$\displaystyle\:C_{2}H_{2}= 30 ml, CO= 26 ml$
0%
$\displaystyle\:C_{2}H_{2}= 33 ml, CO= 28 ml$

Explanation

Let the original mixture contains x ml of CO and (40-x) ml of

$C_2H_2$ .
Th balanced chemical equations are as shown below.

$\underset {40-x}{C_2H_2} + \underset {2.5 (40-x)} {\dfrac {5}{2}O_2} \rightarrow \underset {2(40-x)} {2CO_2}+H_2O(l)$

$\displaystyle\underset {x} CO+ \underset {0.5x}{\dfrac {1} {2}O_2} \rightarrow \underset {x}{CO_2}$
Total volume is 104 ml.
Hence,

$2(40-x)+x+(100-2.5(40-x)-0.5x=104$ .
On solving this, we get

$x=24\ ml$ .
Hence, the volumes of CO and

$C_2H_2$ are 24 ml and

$40-24=16$ ml respectively.

$0.60$ g nitrogen containing compound was boiled with

$NaOH$ , and

$NH_3$ thus formed, required

$100$ mL of

$0.2$ N

$H_2SO_4$ for neutralisation. The percentage of nitrogen in the compound is:

0%
$46.67\%$
0%
$23.34\%$
0%
$60.00\%$
0%
$20.00\%$

Explanation

100 mL of 0.2 N

$H_2SO_4 \equiv$ 0.02 equivalents of ammonia.

Since, basicity of ammonia is 1, moles of ammonia is also equal to 0.02 mol.

Each mole of ammonia has 1 mole of nitrogen atom, so moles of nitrogen atom is also equal to 0.02 .

So, 0.60 g of organic compound contains

$\displaystyle \frac {14 \times 20}{1000}$ g of nitrogen.

Therefore, % of nitrogen in the compound

$= \dfrac{Mass\ of\ nitrogen}{Mass\ of \ compound}\times 100=\displaystyle \frac {14 \times 20 \times 100}{1000 \times 0.6} = 46.67$ % of nitrogen.

So, percentage of nitrogen is

$46.67$ %

Number of

$K^+$ ions and number of moles of

$K^+$ ions present in

$1\:L$ of

$\displaystyle\frac{N}{5}\:KMnO_4$ solution respectively, in acidic medium are :

0%
$0.04$ and $2.4\times{10}^{22}$
0%
$2.4\times{10}^{22}$ and $0.04$
0%
$200$ and $6.023\times{10}^{23}$
0%
$6.023\times{10}^{23}$ and $200$

Explanation

Number of moles in one litre of

$\dfrac{N}{5} KMnO_4$ solution

$= \dfrac{Normality}{n_f} =\dfrac{1}{5\times 5}= \dfrac{1}{25} = 0.04\ mol$ .

So, no. of moles of

$K^+ = 0.04\ mol$ and number of

$K^+ = N_A \times 0.04 = 2.4 \times {10}^{22}$ .

Which of the following have equal mass of

$Cl^-$ ions in

$1.0$ L of each of the following solutions?

0%
$5\%\ NaCl$ (density $= 1.07$ g/mL)
0%
$5\%\ KCl$ (density $=1.06$ g/mL)
0%
$58.5$ g of $NaCl$
0%
$55.5$ g of $BaCl_2$

Explanation

Mass of

$Cl^{\ominus}$ is 35.5grams
Use,

$M =\dfrac {\text {percentage by weight} \times 10\times d}{Mw_2}$
a.

$M=\dfrac {5\times 10\times 1.07}{58.5}=0.91 M (Cl^{\ominus}=0.91, M=0.91\times 35.5=32.46 g)$
b.

$M=\dfrac {5\times 10\times 1.06}{74.5}=0.77 M (Cl^{\ominus}=0.77, M=0.77\times 35.5=25.25 g)$ .

$M=\dfrac {W_2\times 100}{Mw_2\times V_{sol}(in mL)}$

$=\dfrac {58.5\times 1000}{58.5\times 1000}=1 M$

$(Cl^{\ominus}=1 M=1\times 35.5=35.5 g)$
d.

$M=\dfrac {55.5\times 1000}{111\times 1000}=0.5 M$

$BaCl_2\rightarrow Ba^{2+}+2Cl^{\ominus}$

$\therefore Cl^{\ominus}=2\times M=2\times 0.5=1 M=1\times 35.5=35.5 g$
Hence options C and D are correct.

The

$\%$ loss in weight after heating a pure sample of potassium chlorate (Molecular weight

$= 122.5$ g/mol) will be:

0%
$12.25$
0%
$24.5$
0%
$39.2$
0%
$19.6$

Explanation

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

$\dfrac {96}{245} \times 100=39.2$ %.

The number of

${Ag}^{+}$ ion in one drop of silver nitrate solution is:

0%
$2.12\times {10}^{20}$
0%
$7.2\times {10}^{20}$
0%
$7.2\times {10}^{21}$
0%
$7.2\times {10}^{19}$

Explanation

$1$ drop of

$AgN{O}_{3} =12.5\times {10}^{-3}$ g

${Cl}^{-}$ (Given)

$1$ drop of

$AgN{O}_{3} \equiv \displaystyle\frac{12.5\times {10}^{-3}}{35.5}$ mol

${Cl}^{-}$

$=3.52\times {10}^{-4}$ mol

${Cl}^{-}$

$=3.52\times {10}^{-4}$ mol

$AgN{O}_{3}$

$=3.52\times {10}^{-4}$ mol

${Ag}^{+}$

$\equiv 3.52\times {10}^{-4}\times 6.023\times {10}^{23}$ ions of

${Ag}^{+}$ (as each mole has

$N_A$ molecules)

$=2.12\times {10}^{20}$ ions of

${Ag}^{+}$

Hence, the correct option is

$\text{A}$

If the percentage of water of crystallization in

$MgSO_4.xH_2O$ is

$13\%$ , then what is the value of

$x$ ?

0%
$1$
0%
$4$
0%
$5$
0%
$7$

Explanation

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

$A$

The percentage of

${P}_{2}{O}_{5}$ in diammonium hydrogen phosphate

$[{(N{H}_{4})}_{2}HP{O}_{4}]$ is :

0%
$23.48 \%$
0%
$53.78 \%$
0%
$46.96 \%$
0%
$71.0 \%$

Explanation

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

$53.78$ %

Mole fraction of ethanol in ethanol-water mixture is

$0.25$ . Hence, the percentage concentration of ethanol by weight of the mixture is:

0%
$25\%$
0%
$75\%$
0%
$46\%$
0%
$54\%$

Explanation

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

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

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

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

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

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

Therefore, % concentration of ethanol by weight in mixture

$=46$ %.

Hence, the correct option is

$C$

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 :

0%
$65$
0%
$25$
0%
$75$
0%
$35$

Explanation

$MCO_3\rightarrow MO+CO_2\uparrow$

$M'CO_3\rightarrow M'O+CO_2\uparrow$
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

$= 100-50-15=35$ %

A hydrate of

$Na_2SO_3$ has

$50\%$ water by mass. It is :

0%
$Na_2SO_3\cdot 4H_2O$
0%
$Na_2SO_3\cdot 6H_2O$
0%
$Na_2SO_3\cdot 7H_2O$
0%
$Na_2SO_3\cdot 2H_2O$

Explanation

$Na_2SO_3:H_2O\equiv 50:50$ 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

$Na_2SO_4.7H_2O$ .

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 :

0%
$4.34$
0%
$2.17$
0%
$5.28$
0%
$8.34$

Explanation

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.

$1000$ g sample of

$NaOH$ contains

$3$ moles of

$O$ atom, what is the purity of

$NaOH$ ?

0%
$14\%$
0%
$100\%$
0%
$12\%$
0%
$24\%$

Explanation

$1000$ 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

$C$

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:

0%
$90$
0%
$49$
0%
$30$
0%
$47$

Explanation

$1$ 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

$\dfrac {882}{882+910}\times 100 = 49.2$ %.

$Na_2SO_\:xH_2O$ has

$50\%$

$H_2O$ . Hence,

$x$ is :

0%
$4$
0%
$5$
0%
$6$
0%
$8$

Explanation

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,

$\displaystyle \dfrac {142+18x}{2}=18x$

$36x=142+18x$

$18x=142$

$x=7.9 \simeq 8$

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:

0%
$176.5$
0%
$252.2$
0%
$287.6$
0%
$360.1$

Explanation

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.

The sodium salt of methyl orange has

$7\%$ sodium. What is the minimum molecular weight of the compound?

0%
$420$
0%
$375$
0%
$329$
0%
$295$

Explanation

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

$C$

Mole fraction of ethanol and water mixture is

$0.25$ . Hence, percentage concentration of ethanol by weight of mixture is:

0%
$25\%$
0%
$75\%$
0%
$46\%$
0%
$54\%$

Explanation

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 $C$ .

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:

0%
$XY$
0%
$X_2Y$
0%
$XY_2$
0%
$X_2Y_3$

Explanation

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 $B$ .

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 :

0%
$10$
0%
$15$
0%
$20$
0%
$25$

Explanation

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

$Fe$ .

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)

0%
$18$ g of water is formed by the combination of $16$ g oxygen with $2$ g of hydrogen.
0%
$18$ g of water in liquid state is obtained by heating $18$ g of ice.
0%
$18$ g of water is completely converted into vapour state on heating.
0%
$18$ g of water freezes at $4^oC$ to give same mass of ice.

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?

0%
$74.4$
0%
$70$
0%
$80$
0%
$90$

Explanation

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 %

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.

0%
$\;25\%$
0%
$\;10\%$
0%
$\;18\%$
0%
$\;15.5\%$

Explanation

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

$\displaystyle = \frac {18}{116.2} \times 100 =15.5$ %

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.

0%
$\displaystyle 83.5 + \frac{x}{4}$
0%
99.5 + $x$
0%
83.5 + $x$
0%
$\displaystyle 99.5 + \frac{x}{2}$

Explanation

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

$\text{B}$ .

If 40 g of ethyl alcohol is dissolved in 50 ml of water, then calculate the weight/volume percentage of ethyl alcohol present in the solution? [Density of ethyl alcohol = 0.8 g/ml].

0%
40%
0%
30%
0%
25%
0%
35%

Explanation

Weight/ Volume of component

$= \cfrac {Mass \ of \ component}{Total \ volume \ of \ solution} \times 100$

Mass of component i.e ethyl alcohol

$=40 \ g$

Volume of water

$=50 \ ml$

Volume of ethyl alcohol

$= \cfrac {mass}{density}= \cfrac {40}{0.8}=50 \ ml$

Total volume of solution

$= Volume \ of \ H_2O + Volume \ of \ ethyl \ alcohol$

$=50+50=100 \ ml$

W/V of component

$= \cfrac {40}{100} \times 100 = 40$ %

$\displaystyle n_{1}$ 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 :

0%
$\displaystyle n_{1}-n_{2}=m_{1}-m_{2}$
0%
$\displaystyle n_{1}+n_{2}=m_{1}+m_{2}$
0%
$\displaystyle n_{1}=n_{2}$
0%
$\displaystyle m_{1}=m_{2}$

Explanation

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.

A certain compound has the molecular formula,

$\displaystyle X_{4}O_{6}$ . If

$10$ g of compound contains

$5.62$ g of X, the atomic mass of X is approximately :

0%
32 amu
0%
30.8 amu
0%
42 amu
0%
48 amu

Explanation

10 g of

$X_4O_6$ contains

$5.62$ g of X.

So, the weight of

$O$ is

$10 - 5.62 = 4.38\:g$ .

Now, as

$\dfrac{weight}{atomic\: mass} =$ number of atoms in given formula,

So,

$\dfrac{Weight\: of\: X}{atomic\: mass\: of\: X}$ :

$\dfrac{Weight\: of\: O}{atomic\: mass\: of\: O}= 4:6$

Let atomic mass of X be

$x$ .

$\implies \dfrac{5.62}{x}$ :

$\dfrac{4.38}{16} = 4:6$

$\implies \dfrac{5.62\times 16}{4.38x} = \dfrac{4}{6}$

$\implies \dfrac{5.62\times 16\times 6}{4} = 4.38x$

So,

$x = 30.8 \: amu$ .

The number of electrons required to deposit

$1$ g atom of Al (atomic wt.

$=27$ ) from a solution of

$AlCl_3$ is:

0%
$1N_A$
0%
$2N_A$
0%
$3N_A$
0%
$4N_A$

Explanation

1 g atom equals to 1 mole.

$AlCl_3$ solution decomposes to

$Al^{3+}$ and

$3Cl^-$ .

To obtain

$Al$ from

$Al^{3+}$ , 3 mole of electrons are required.

$Moles=\dfrac{number\:of\:atoms}{N_A}$

Number of electrons required=

$3N_A$

The hydrated salt,

$\displaystyle Na_{2}SO_{4}.nH_{2}O$ undergoes 56% loss in weight on heating and becomes anhydrous. The value of n (approx.) will be :

0%
5
0%
3
0%
7
0%
10

Explanation

100 g of

$\displaystyle Na_{2}SO_{4}.nH_{2}O$ contians 56 g of water and 44 g of

$\displaystyle Na_{2}SO_{4}$ .

The molar mass of

$\displaystyle Na_{2}SO_{4}$ is

$23+23+32+4(16) = 142 \: g$ .

44 g of

$\displaystyle Na_{2}SO_{4}$ corresponds to 56 g of water.

142 g of

$\displaystyle Na_{2}SO_{4}$ will correspond to

$\displaystyle \frac {142 \times 56}{44}=180$ g of water.

This corresponds to

$\displaystyle \frac {180}{18}=10$ moles.

Thus, the value of n is 10.

Hence, option D is correct.

For 7 molal

$NaOH$ solution. Select the correct statement

0%
% $\left(\displaystyle\frac{w}{w}\right) = 28$
0%
% $\left(\displaystyle\frac{w}{w}\right) = 72$
0%
${ X }_{ { H }_{ 2 }O }= \displaystyle\frac { 7 }{ 47 }$
0%
${ X }_{ NaOH }= \displaystyle\frac { 7 }{ 47 }$

Explanation

1 molal NaOH solution means 1 moles of solute dissolved in

$1000$ gms of solvent

Therefore, number of moles = 7

moles = weight / 40

weight = 280 gms

%(w/w) = weight of solute in grams x 100 / weight of solutions in gms

%(w/w) =

$28$

"All gases have the same number of moles in the same volume at constant temperature and pressure:. This statments belongs to :

0%
Boyle's law
0%
Charles's law
0%
Avogadro's principle
0%
ideal gas law
0%
Dalton's law

Statement 1: At the same temperature and pressure, 1 L of hydrogen gas and 1 L of neon gas have the same mass.
Statement 2: Equal volumes of ideal gases at the same temperature and pressure contain the same number of moles.

0%

Both statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1
0%

Both statement 1 and statement 2 are correct but Statement 2 is not the correct explanation of Statement 1
0%

Statement 1 is correct but statement 2 is incorrect
0%

Statement 1 is incorrect but statement 2 is correct
0%

Both the statement 1 and statement 2 are incorrect

CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 12 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 12 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.