Solutions Test 40

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Solutions Test 40

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Question 1
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Directions For Questions

Vitamin C $$(M=176)$$ is a compound of $$C, H$$ and $$O$$ found in many natural sources, especially citrus fruits. When a $$1.0\ g$$ sample of vitamin C is placed a combustion chamber and nurned, the following data are obtained:
Mass of $$CO_{2}$$ absorber after combustion $$=85.35\ g$$
Mass of $$CO_{2}$$ absorber after combustion $$=83.85\ g$$
Mass of $$H_{2}O$$ absorber after combustion $$=37.96\ g$$
Mass of $$H_{2}O$$ absorber after combustion $$=37.55\ g$$

...view full instructions

What is the percentage of carbon, by wt. in vitamin $$C$$ ?

A
$$66.67\%$$

B
`$$40.9\%$$

C
$$20\%$$

D
$$60\%$$

Solution

Mass of $$CO_{2}=85.35-83.85=1.5\ g$$
Mass of $$H_{2}O=37.96-37.55=0.41\ g$$
Mole of $$CO_{2}\dfrac{1.5}{44}=0.034\ g$$
Mass pf $$C=0.034\times 12=0.41\ g$$
Mole of $$H_{2}O=\dfrac{0.41}{13}=0.0227$$
Mole of $$H=0.0227\times 2=0.0455\ mole$$
Mass fo $$H=0.0455\times 1=0.0455\ g$$
Mass of oxygen $$=1- (0.41+0.0455)=0.544\ g$$
$$\% C=\dfrac{0.41}{18}\times 100=41\%$$ Option $$(b)$$
Question 2
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Directions For Questions

Figure shows a scheme, for concentrating an dilute solution of $$NaOH$$

...view full instructions

How much concentrated solution obtained per hour?

A
$$5000\ kg$$

B
$$500\ kg$$

C
$$19,500\ kg$$

D
$$20,000\ kg$$

Solution

$$2\%NaOH=\dfrac{x}{25000}\times 100$$
$$x=\dfrac{2\times 25000}{100}=500\ kg\ NaOH$$ & water
$$18\%=NaCl=\dfrac{x}{25000}\times 100$$
$$x=\dfrac{18\times 25000}{100}=4500\ kg\ NaCl$$
$$90\% NaCl=\dfrac{4500}{y}\times 100$$
$$y=\dfrac{4500}{90}\times 100=5000$$
$$5\%=$$ water on slurry & Care soln $$=500$$
man of mole in feed streak $$=0.8\times 2.500=20000$$
water evaporated $$=20,000-500=19,500$$
Concentrated soln $$=x=$$ max of $$NaOH+$$ water $$=500\ kg$$ Option $$(b)$$
Question 3
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1.0 g of moist sample of a mixture of $$KCl$$ and $$KClO_3$$ was dissolved in water and made up to 250mL. 25 mL of this solution was treated with $$SO_2$$. The chlorate was reduced to chloride and an excess of $$SO_2$$ was removed by boiling. The total chloride was precipitated as $$AgCl$$ weighing 0.1435 g. In another experiment, 25 mL of the original solution was heated with 30 mL of 0.2 N solution of ferrous sulphate and the unreacted ferrous sulphate required 37.5 mL of 0.08 N solution of an oxidising agent for complete oxidation. For the given reaction:

$$ClO_3^{ɵ} +6Fe^{2+}+6H^{+} \rightarrow Cl^{ɵ}+6Fe^{3+}+3H_2O$$

What is the mass per cent of potassium chloride in the moist sample?

A
37.25%

B
61.25%

C
3.725%

D
74.5%
Question 4
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To determine soluble (free) $$SiO_2$$ in a rock, an alkaline extraction was carried out, as a result of which there was found $$1.52\%$$ of $$SiO_2$$ in the extract and also $$1.02\%$$ of $$Al_2O_3$$. Considering that, apart from the free $$SiO_2$$, the extract also contained the $$SiO_2$$, that had passed into it from Kaolin $$(2SiO_2, Al_2O_3)$$, the percentage of $$SiO_2$$, in the rock being analysed is $$(Si=28, Al=27)$$

A
$$1.20$$

B
$$0.32$$

C
$$0.50$$

D
$$1.52$$

Solution

Total $$\%$$ of $$SiO_2$$observed $$=1.52\%$$
$$\%$$ of $$Al_2O_3=1.02\%$$
$$\%$$ of $$SiO_2=$$ out of $$SiO_2$$ from $$2SiO_2.Al_2O_3$$
And $$Al_2O_3$$ is from only $$2SiO_2.Al_2O_3$$
Mass of $$SiO_2$$ in $$2SiO_2.Al_2O_3=2\times (38+32)$$
$$=120\ gm$$
Mass of $$Al_2O_3$$ in $$2SiO_2.Al_2O_3 =27\times 2+16\times 3$$
$$=102\ gm$$
Now $$102\ gm\ Al_2O_3$$ contribute $$102\%$$
$$\therefore \ 1\ gm$$ contribute $$0.01\%=\dfrac {1.02}{102}\%$$
$$\therefore \ 120\ gm\ SiO_2.Al_2O_3$$ contribute
$$120\times 0.01=1.20\%$$
$$\therefore \ \%$$ of $$SiO_2$$ in nock $$=1.52-1.20$$
$$=0.32\%$$
Question 5
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Directions For Questions

Figure shows a scheme, for concentrating an dilute solution of $$NaOH$$

...view full instructions

How much slurry obtained per hour?

A
$$5000\ kg$$

B
$$500\ kg$$

C
$$19,500\ kg$$

D
$$20,000\ kg$$

Solution

$$2\%NaOH=\dfrac{x}{25000}\times 100$$
$$x=\dfrac{2\times 25000}{100}=500\ kg\ NaOH$$ & water
$$18\%=NaCl=\dfrac{x}{25000}\times 100$$
$$x=\dfrac{18\times 25000}{100}=4500\ kg\ NaCl$$
$$90\% NaCl=\dfrac{4500}{y}\times 100$$
$$y=\dfrac{4500}{90}\times 100=5000$$
$$5\%=$$ water on slurry & Care soln $$=500$$
man of mole in feed streak $$=0.8\times 2.500=20000$$
water evaporated $$=20,000-500=19,500$$
Concentrated soln $$=x=$$ max of $$NaOH+$$ water $$=500\ kg$$
Mass of slurry $$=y=5000\ kg$$ Option $$(a)$$
Question 6
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In $$200\ g$$ of a sample of olem labelled as $$109.0\% ,\ 12\ g$$ water is added. The new labelling of the oleum sample is

A
$$106.0\%$$

B
$$103.0\%$$

C
$$102.8\%$$

D
$$105.6\%$$

Solution

Given: $$200\ gm$$ of sample of oleum
labelled $$109\%$$
$$12\ gm\ H_2O $$ is added
$$503+H_2O\to H_2SO_4$$
$$80\ gm\ 18\ gm\ 98\ gm$$
$$109\%$$ means $$100\ gm$$ sample of okeam $$(SO_3+H_2SO_4)$$ in which $$9\ gm\ H_2O$$ react with $$\% SO_3$$
$$18\ gm\ H_2O\to 80\ gm\ SO_3$$
$$\therefore \ 9\ gm\ H_2O\to \dfrac {9}{18}\times 80=40\ gm\ SO_3$$
in $$100\ gm$$ sample
$$\therefore \ wt$$ of $$H_2SO_4=60\ gm$$ in $$100\ gm$$ sample
Now, $$12\ gm\ H_2O$$ is added
$$18\ gm\ H_2O\to 80\ gm\ SO_3$$
$$\therefore \ 12\ g\ H_2O\to \dfrac {12}{18}\times 80$$
$$=53.33\ gm\ SO_3$$ reacted to from $$(53.33+12)\ gm\ H_2SO_4=65.33\ gm\ H_2SO_4$$
Now,
In $$200\ g$$ sample :- $$SO_3=80\ gm$$
$$H_2SO_4=120\ gm$$
After adding $$H_2O; \ SO_3=80--53.33\ gm =26.67\ gm$$
$$H_2SO_4=120+65.33\ gm$$
$$=185.33\ gm$$
$$\%$$ of $$SO_3=\dfrac {26.67}{212}\times 100=12.67$$
$$\therefore \ 12.6\ gm\ SO_3$$ react with:- $$2.8\ gm\ H_2O$$
$$\therefore \ \%$$ label on sample $$=100+2.8=102.8\%$$
Question 7
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When $$200\ g$$ of an oleum sample labelled as $$109\%$$ is mixed with $$300\ g$$ of another oleum sample labelled as $$118\%$$ the new labelling of resulting oleum sample becomes

A
$$114.4\%$$

B
$$112.6\%$$

C
$$113.5\%$$

D
$$127\%$$

Solution

$$200\ gm$$ of $$109\%$$ oleum sample
$$300\ gm$$ of $$118\%$$ oleum sample
$$SO_3+H_2O\to H_2SO_4$$
$$80\ gm\ 18\ gm\ 28\ g$$
In $$109\%$$ labelled sample
$$9\ gm\ H_2O$$ react with $$\dfrac {80}{18}\times 9=40\ gm\ SO_3$$
$$\therefore \ 40\%\ SO_3+60\% \ H_2SO_4$$
$$\therefore \ $$ In $$200\ gm$$ sample $$80\ gm\ SO_3$$
$$12\ gm\ H_2SO_4$$
Now
In $$118\%$$ sample
$$18\ gm\ H_2O$$ react with $$80\ gm\ SO_3$$
$$\therefore \ 80\% \ SO_3\ ;\ 20\% \ H_2SO_4$$
$$\therefore \ $$ In $$300\ gm$$ sample :- $$240\ gm\ SO_3$$
$$60\ gm\ H_2SO_4$$
When both sample mined:-
$$SO_3 :- 80+240=320\ gm\ H_2SO_4 :- 120+60=180\ gm$$
$$\therefore \ \%$$ of $$SO_3=\dfrac {320}{320+180}\times 100=64\%$$
$$64\ gm\ SO_3$$ combine with :- $$\dfrac {18}{80}\times 64=14.4\ gm\ H_2O$$
$$\therefore \ $$ New label of resulting sample
$$=100+14.4$$
$$=114.4\%$$
Question 8
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A sample of oleum is labelled as $$112\%$$. In $$200\ g$$ of this sample, $$18\ g$$ water is added, The resulting solution will contain

A
$$218\ g$$ pure $$H_2SO_4$$

B
$$218\ g\ H_2SO_4$$ and $$6\ g$$ free $$SO_3$$

C
$$212\ g\ H_2SO_4$$ and $$6\ g$$ free $$SO_3$$

D
$$191.33\ g\ H_2SO_4$$ and $$26.67\ g$$ free $$SO_3$$

Solution

$$200\ g$$ sample of oleum labelled $$112\%$$
$$SO_3+H_2O\to H_2SO_4$$
$$80\ gm\ 18\ gm\ 98\ gm$$
$$112\%$$ sample
$$12\ gm\ H_2O$$ react with $$\dfrac {80}{18}\times 12$$
$$=53.33\ gm\ SO_3$$
$$\therefore \ \% SO_3=53.33\%$$ (free $$SO_3$$)
$$\%\ H_2SO_4=46.67\%$$
In $$200\ mg$$ sample
$$SO_3=106.66\ gm$$
$$H_2SO_4=93.34\ gm$$
Now $$18\ g\ H_2O$$ is added
$$\therefore \ SO_3$$ converted to $$H_2SO_4=80\ gm$$
and from $$98\ gm\ H_2SO_4$$
Now $$SO_3=106.66-80=26.66\ gm\ SO_3$$
$$H_2SO_4=93.34+98=191.34\ gm\ H_2SO_4$$
Question 9
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Vapour pressure of liquid

A
increases with increase in temperature

B
decrease with increase in temperatuyre

C
is independent of temperature

D
either increases or decreases with the increase in temperature, depending on the nature of liquid

Solution
Question 10
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An aqueous solution of sucrose is 0.5 molal. What is the vapour pressure of water above this solution ? The vapour pressure of pure water is 25.0 mm Hg at this temperature.

A
24.8 mm Hg

B
0.45 mm Hg

C
2.22 mm Hg

D
20.3 mm Hg

Solution

Given that
$$ m = 0.5\;\;molal $$
V.P. (pure water) 25.0mm Hg

According to the Raoult's law for dilute solution
$$\dfrac{P_0 - P_s}{P_0} = \dfrac{w}{m+W}\times 1000 = X_{sucrose}$$

here the 0.5 molal of solution means 0.5 mole in 1kg of water
Henc the mole of solvent = $$\dfrac{1000}{18}=55.5$$ [ the molecular mass of water molecules is 18]

then $$ \dfrac{25-P_s}{25} = \dfrac{0.5}{55.57+0.5}$$

$$ \Rightarrow\;1401.75 - 56.07P_s = 12.5 $$

$$\Rightarrow \; 56.07P_s = 1407.75 - 12.5$$

$$\Rightarrow \; P_S = \dfrac{1389.25}{\;\;56.07}$$

$$24.8\;mm \;Hg$$

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Solutions Test 40

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Solutions Test 40

Score

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SHARING IS CARING

Question 1

$$66.67\%$$

`$$40.9\%$$

$$20\%$$

$$60\%$$

Solution

Question 2

$$5000\ kg$$

$$500\ kg$$

$$19,500\ kg$$

$$20,000\ kg$$

Solution

Question 3

37.25%

61.25%

3.725%

74.5%

Question 4

$$1.20$$

$$0.32$$

$$0.50$$

$$1.52$$

Solution

Question 5

$$5000\ kg$$

$$500\ kg$$

$$19,500\ kg$$

$$20,000\ kg$$

Solution

Question 6

$$106.0\%$$

$$103.0\%$$

$$102.8\%$$

$$105.6\%$$

Solution

Question 7

$$114.4\%$$

$$112.6\%$$

$$113.5\%$$

$$127\%$$

Solution

Question 8

$$218\ g$$ pure $$H_2SO_4$$

$$218\ g\ H_2SO_4$$ and $$6\ g$$ free $$SO_3$$

$$212\ g\ H_2SO_4$$ and $$6\ g$$ free $$SO_3$$

$$191.33\ g\ H_2SO_4$$ and $$26.67\ g$$ free $$SO_3$$

Solution

Question 9

increases with increase in temperature

decrease with increase in temperatuyre

is independent of temperature

either increases or decreases with the increase in temperature, depending on the nature of liquid

Solution

Question 10

24.8 mm Hg

0.45 mm Hg

2.22 mm Hg

20.3 mm Hg

Solution

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