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CBSE Questions for Class 12 Engineering Chemistry Solutions Quiz 15 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Solutions
Quiz 15
The amount of anhydrous $$Na_{2}CO_{3}$$ present in 250 ml of 0.25 M solution is
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6.225 g
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66.25 g
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6.0 g
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6.625 g
When NaCl is dissolved in water
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boiling point of solution decreases
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boiling point of solution increases
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boiling point of solution remains unchanged
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freezing point of solution increases
Moles of $${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$$ to be dissolved in 12 moles of water to lower its vapour pressure by 10mm Hg at a temperature at which vapour pressure of pure water is 50 mm Hg is:
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1.5 moles
0%
2 moles
0%
1 mole
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3 moles
Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y?
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0%
20
0%
10
0%
100
0%
30
When solute and solvent both are volatile, the vapour pressure of solution is directly proportional to mole fraction of which?
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Solute
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Solvent
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Solute and Solvent
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none
Select the correct statement(s) from the following?
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Addition of $$Hg{I_2}$$ to aqueous solution of $$KI$$ increases the freezing point since a complex $${K_2}\left[ {Hg{I_4}} \right]$$ is formed
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Three aqueous solutions of $$NH_3$$ labelled as A, B and C with concentration 0.1 M, 0.01 M and 0.001 M respectively, then the value of Van't Hoff factor for these solution will be in order, $${i_A} < {i_B} < {i_C}$$ (consider degree of dilution)
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Vapour density of $$PC{l_5}\left( g \right)$$ dissociating into $$PC{l_3}\left( g \right)$$ and $$C{l_2}\left( g \right)$$ is 100, then the Van't Hoff factor will be $$4.25$$
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If $$pH$$ of $$0.1 M$$ monobasic acid is found to be 2, its osmotic pressure at a given temperature $$T$$ kelvin will be $$0.11 RT$$
Four solutions of $$K_2SO_4$$ with the following concentration $$0.1 m$$, $$0.01 m$$, $$0.001 m$$ and $$0.0001 m$$ are available. The maximum value of van't Hoff factor, i, corresponds to:
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0.0001 m solution
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0.001 m solution
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0.01 m solution
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0.1 m solution
The vapour pressure of $$CCl_4$$ at $$25^0C$$ is 143 mm of Hg
. 0.5 gm of a non-volatile solute(mol. wt.
65
) is dissolved in
100ml
of $$CCl_4$$,
the vapour pressure of the solution will be:
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199.34 mm Hg
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143.99 mm Hg
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141.43 mm Hg
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94.39 mm Hg
$$0.6g$$ urea is added to $$360g$$ water. Calculate lowering in vapor pressure for this solution
(Given: Vapour pressure of $$H_2O$$ is $$35mm$$ of $$Hg$$)
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$$0.027mm$$ of $$Hg$$
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$$0.035mm$$ of $$Hg$$
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$$0.017mm$$ of $$Hg$$
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$$0.040mm$$ of $$Hg$$
Explanation
Solution:- (C) $$0.017mm$$ of $$Hg$$
$$\dfrac{P^0-Ps}{P^0}=\dfrac{n}{n+N}$$
Lowering in V.P.= $$P^0\times \dfrac{n}{n+N}$$
$$=35\times \dfrac{\dfrac{0.6}{60}}{\dfrac{0.6}{60}+\dfrac{360}{18}}=0.017mm$$ of Hg
In a mixture of $$A$$ and $$B$$, having vapour pressure of pure $$A$$ and pure $$B$$ as $$400mm\, Hg$$ and $$600mm\, Hg$$ respectively, mole fraction of $$B$$ in liquid phases is $$0.5$$. Calculate total vapour pressure and mole fraction of $$A$$ and $$B$$ in vapour phases.
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$$500, 0.4, 0.6$$
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$$500, 0.5, 0.5$$
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$$450, 0.4, 0.6$$
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$$450, 0.5, 0.5$$
Explanation
$$P_T = X_AP_{A^o} + X_BP_{B^o}$$
$$= 0.5\times 400+0.5\times 600$$
$$=500nm$$ of $$Hg$$
$$\dfrac{1}{P_T} = \dfrac{y_A}{P^o_A} + \dfrac{1-y_B}{P^o_B}$$
$$y_A = 0.6$$ & $$y_B = 1 -0.6 = 0.4$$
When a substance is dissolved in a solvent, the vapour pressure of solvent decrease. It brings
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a decrcase in boilings point of solution
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an increase in boiling point of the solution
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a decrease in freezing point of the solution
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an increase in freezeing point of the solution
Explanation
As we dissolve some substance in the solvent ,it vapour pressure decreases as the mixture gets impure due to this solute.
Thus boiling point also increases. and freezing point decreases
A sample of pure Cu $$(3.18\,g)$$ when heated in the presence of oxygen forms a black oxide of copper $$(CuO)$$. The final weight of CuO is $$3.92$$g. The percent of Cu oxidised is
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$$65\%$$
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$$69\%$$
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$$76\%$$
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$$98\%$$
One mol of a solute A is dissolved in a given volume of a solvent. The association of the solute take place according to $$n A = (A)_n$$. The van't Hoff factor i is expressed as
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$$i = 1 - x$$
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$$i = 1 + \dfrac{x}{n}$$
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$$i = \dfrac{i - x + \dfrac{x}{n}}{1}$$
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$$i = 1$$
How many types of solutions are possible by the combination of any two of three states of matter (solid, liquid, and gas)
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$$3$$
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$$6$$
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$$9$$
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$$18$$
What is the percentage of hydrogen by wt.in vitamin $$C$$ ?
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$$4.55\%$$
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$$41\%$$
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$$20.5\%$$
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$$9.11\%$$
Explanation
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 $$=1- (0.41+0.0455)=0.544\ g$$
$$\% C=\dfrac{0.41}{18}\times 100=41\%$$
$$\%H =\dfrac{0.0455}{1}\times 100=4.55\%$$ Option $$(a)$$
What is the percentage of carbon, by wt. in vitamin $$C$$ ?
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$$66.67\%$$
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`$$40.9\%$$
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$$20\%$$
0%
$$60\%$$
Explanation
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 $$(b)$$
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?
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37.25%
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61.25%
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3.725%
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74.5%
What is the percentage of silica present in the ore, by weight?
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$$2.82$$
0%
$$3.02$$
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$$0.465$$
0%
$$6.05$$
Explanation
Mole wt of kaolin $$=2589/mole$$
Mole of $$Al$$ is kaolin $$=\dfrac{13}{258}=0=5$$
$$(Al_{2}O_{3})$$
Mole wt of Gibbrite $$=156 9/mole$$
Mole of gibbrite $$=\dfrac{87}{156}=0.557\ mole$$
Total mole of $$Al_{2}O_{3}=0.557+0.05=0.607\ mole$$
Mole of $$HO_{2}$$ from kaolin $$=2\times 0.05=0.01\ mole$$
Max mole of Muld forms $$=\dfrac{0.1}{5}=0.02\ mole$$
$$1$$ mole Mud $$=3$$ mole $$N_{2}O_{3}$$
$$0.02$$ mole Mud $$=x-1$$
$$x=3\times 0.02=0.06$$
Max $$Al_{2}$$ can be expected $$=0.607-0.06=0.547$$
$$\% Al$$ can be expected $$=\dfrac{0.547}{0.607}\times 100=90.11\%$$
mole $$SiO_{2}=0.1\ mole$$
Max $$SiO_{2}=0.1\times 60=6g$$
$$\%$$ sodia $$=\dfrac{6}{15}\times 100=6\%$$ Option $$(d)$$
Calculate van't Hoff factor for $$0.2\ m$$ aqueous solution of $$KCl$$ which freezes at $$-0.680^o$$C. ($$K_f=1.86$$ K kg $$mol^{-1}$$)
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$$3.72$$
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$$1.83$$
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$$6.8$$
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$$1.86$$
Explanation
Given,
$$m =0.2m$$
$$K_f=1.86\ K\ Kg\ mol^{-1}$$
$$T_f=-0.680^0C$$
We have,
$$(\Delta T_f)$$ $$=i\times K_f \times m$$
Here, $$i=$$ Van't Hoff factor
$$\Delta T_f=0-0.680=0.680^0C$$
Here, $$i=\dfrac{\Delta T_f}{K_f\times m}=\dfrac{0.68}{1.86\times 0.2}=1.83$$
Hence, option $$B$$ is correct.
A $$1.50\ g$$ sample of type metal (an alloy of $$Sn, pb, Cu$$ and $$Sb$$) is dissolved in nitric acis and metastannic acid, $$H_2SnO_3$$, prpitates. This is delydrated by heating to tin $$(IV)$$ oxide, which is found to weight $$0.50\ g$$. What percentage of tin was in the original type metal sample? $$(Sn=119)$$
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$$33.33\%$$
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$$26.27\%$$
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$$29.38\%$$
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$$52.54\%$$
Explanation
weight of $$SnO_2=0.50\ g$$
$$151\ g\ SnO_2=119\ g\ Sn$$
$$0.50\ SnO_2=\dfrac {119\times 0.50}{151} $$
$$=0.3940\ g\ Sn$$
$$\%$$ of $$Sn=\dfrac {0.3940}{1.50}\times 100$$
$$=26.269\%$$
$$\%$$ of $$Sn=26.27\%$$
When the air temperature is below freezing, the saturation vapor pressure over water is _______.
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Equal to zero
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Less than the saturation vapor pressure over ice
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Greater than the saturation vapor pressure over ice
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Equal to the saturation vapor pressure over ice
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 $$(Si=28, Al=27)$$
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$$1.20$$
0%
$$0.32$$
0%
$$0.50$$
0%
$$1.52$$
Explanation
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 $$=1.52-1.20$$
$$=0.32\%$$
How much slurry obtained per hour?
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$$5000\ kg$$
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$$500\ kg$$
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$$19,500\ kg$$
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$$20,000\ kg$$
Explanation
$$2\%NaOH=\dfrac{x}{25000}\times 100$$
$$x=\dfrac{2\times 25000}{100}=500\ kg\ NaOH$$ & 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 $$(a)$$
How much water is evaporated per hour?
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$$500\ kg$$
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$$500\ kg$$
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$$19,500\ kg$$
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$$20, 000\ kg$$
Explanation
$$2\%NaOH=\dfrac{x}{25000}\times 100$$
$$x=\dfrac{2\times 25000}{100}=500\ kg\ NaOH$$ & 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 $$=20,000-500=19,500$$ Option $$(c)$$
How much concentrated solution obtained per hour?
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$$5000\ kg$$
0%
$$500\ kg$$
0%
$$19,500\ kg$$
0%
$$20,000\ kg$$
Explanation
$$2\%NaOH=\dfrac{x}{25000}\times 100$$
$$x=\dfrac{2\times 25000}{100}=500\ kg\ NaOH$$ & 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 $$(b)$$
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
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$$106.0\%$$
0%
$$103.0\%$$
0%
$$102.8\%$$
0%
$$105.6\%$$
Explanation
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 $$=100+2.8=102.8\%$$
Directions: In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice:
Assertion: Unseeing of vapour pressure is not dependent on the number of species present in the solution.
Reason : Lowering of vapour pressure and relative lowering of vapour pressure are colligative properties.
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If both assertion and reason are true and reason is the correct explanation of assertion.
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If both assertion and reason are true but reason is not the correct explanation of assertion.
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If assertion is true but reason is false.
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If both assertion and reason are false.
Explanation
Lowering of vapour pressure is directly proportional to the number of species present in the solution. Only relative lowering of vapour pressure is a colligative property.
Directions: In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice:
Assertion: The vapour pressure of an aqueous solution is less than 1.013 bar at 373.15 K.
Reason: Vapour pressure of water is 1.013 bar at 373.15K.
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If both assertion and reason are true and reason is the correct explanation of assertion.
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If both assertion and reason are true but reason is not the correct explanation of assertion
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If assertion is true but reason is false.
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If both assertion and reason are false.
Which of the following is correct for an ideal solution?
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$$\Delta H_{mix} = 0$$
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$$\Delta S_{mix} = 0$$
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$$\Delta G_{mix} = 0$$
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None of these
$$2.0\ g$$ mixture of sodium carbonate and sodium bicarbonate on heating to constant weight gave $$224\ mL$$ of $$CO_{2}$$ at N.T.P.
The $$\%$$ weight of sodium bicarbonate in the mixture is:
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0%
$$50$$
0%
$$54$$
0%
$$80$$
0%
$$84$$
Explanation
On heating the mixture only sodium bicarbonate dissociates. Sodium carbonate does not dissociate due to its stability.
$$2NaHCO_3 + heat \rightarrow Na_2CO_3 + CO_2 + H_2O$$
Let the weight of $$Na_2CO_3$$ in the mixture be $$x$$ g
Weight of $$NaHCO_3$$ in the mixture $$=(2-x)$$ g
No. of moles of $$Na_2CO_3$$ $$=\dfrac {x}{106}$$ moles
No. of moles of $$NaHCO_3$$ $$=\dfrac{(2-x)}{84}$$ moles
From the equation:
2 moles of
$$NaHCO_3$$ give 1 mole of $$CO_2$$ gas
$$\dfrac{(2-x)}{84}$$ moles
of
$$NaHCO_3$$ give
$$\dfrac{(2-x)}{84} \times \dfrac {1}{2} = \dfrac {(2-x)}{168}$$
moles of $$CO_2$$ gas
At N.T.P
22400 ml of $$CO_2$$ gas = 1 mole
224 ml of $$CO_2$$ gas = $$\dfrac{1}{22400} \times 224 = 0.01$$ moles
Now
$$\dfrac {(2-x)}{168} = \dfrac {1}{100}$$
$$\Rightarrow x = 0.32$$ g
Weight of sodium bicarbonate $$=(2-0.32) g = 1.68$$ g
Weight percentage of sodium bicarbonate = $$\dfrac{1.68}{2} \times 100 = 84$$ %
The compound $$Na_2 CO_3 . x H_2O$$ has $$50 \% H_2O$$ by mass. The value of "x" is :
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0%
4
0%
5
0%
6
0%
7
0%
8
Explanation
(i) Molar mass of $$Na_2CO_3 = 106 $$ unit
Since mass % of the water in the compound is 50%.
$$\Rightarrow $$ Mass of the hydrated compound $$= 2\times 106=212g$$
i.e. Mass of water $$=106 u$$ per hydrated compound.
$$\therefore $$ Number of water molecules, $$x=\dfrac{mass}{molecular \ mass}=\dfrac{106}{18}\approx 6$$
Hence option C correct.
The vapour pressure of pure liquid A at $$ 27^0 C $$ is
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0%
60 torr
0%
48 torr
0%
49.26 torr
0%
61.58 torr
If the experimenty is performed aqueous $$ AlCl_3 $$ solution (a= 0.8 ) prepared by dissolving 1 mole of $$ AlCl_3 $$ in 17 mole of water and the decrease in the mass of solution in the experiment is found to be 0.18 g then the increase in the mass of absorbent should be
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0%
0.216 g
0%
0.225 g
0%
0.191 g
0%
1.08 g
Explanation
$$AlCl_3 \rightleftharpoons Al^{3+} +3Cl^{-}$$
$$1-0.8\\=0.2$$ $$0.8$$ $$3\times 0.8=\\2.4$$
Total no. of mole$$=0.2 + 0.8 +2.4 =3.4$$
Now, decrease in mass of solution $$\propto P$$
and increase in mass of absorber $$\propto P^o$$
$$\therefore \cfrac{P^o}{P}=X_2=\cfrac{17}{17+3.4}=\cfrac{5}{6}=\cfrac{0.18}{|\Delta m|_{absorber}}$$
$$\therefore$$ Increase in mass of absorber $$=0.216\ gm$$
At $$80^o C$$, the vapour pressure of pure liquid A is 520 mm Hg and that of pure liquid B is 1000 mm Hg. If a mixture solution of A and B boils at $$80^o C$$and 1 atm pressure, the amount of A in the mixture is :
(1 atm=760 mm Hg)
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0%
34 mol%
0%
48 mol%
0%
50 mol%
0%
52mol%
Sugar in urine sample can be detected by:
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Fehling's solution
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Benedict's solution
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Tollen's reagent
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all the three $$(a), (b)$$ and $$(c)$$
$$50 mL$$ of solution of $$ Na_2CO_3 $$ neutralizes $$49.35 mL$$ of $$4 N \ HCl$$ . the reaction is represented as
$$ CO^{2-}_3 +2H^+ \rightarrow CO_2 +H_2O $$
The density of this $$ Na_2CO_3 $$ solution is 1.25 g $$ mL^{-1} ,$$ the percentage of $$ Na_2CO_3$$ in it is :
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0%
$$47.7$$
0%
$$37.7$$
0%
$$26.7$$
0%
$$16.7$$
Dry air is bubbled successively through (i) a solution, (ii) its solvent, and (iii) through $$CaCl_2$$. The lowering of vapour pressure of the solvent due to the addition of solute is equal to
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0%
$$p^0 \times \dfrac {loss\ in\ wt. of\ solvent}{gain\ in\ wt\ of\ CaCl_2}$$
0%
$$\dfrac {loss\ in\ wt. of\ solvent}{gain\ in\ wt\ of\ CaCl_2}$$
0%
$$\dfrac {loss\ in\ wt. of\ solution}{gain\ in\ wt\ of\ CaCl_2}$$
0%
$$\dfrac {loss\ in\ wt. of\ solute}{gain\ in\ wt\ of\ CaCl_2}$$
The reading of pressure gauge at bubble point is :
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0%
500
0%
600
0%
700
0%
None
Explanation
$$ X_A = 0.75 \quad X_B = 0.05 $$ above 500 mm Hg only liquid phase exists .
$$ P_{bubble} point = X_A P^0_A + X_B P^0_B $$
$$ P_{cqycqfyr fcUnq} = X_A P60_A + X_B P^0_B $$
$$ 0.75 +400 + 0.25 \times 800 = 500 mm $$
$$ y_A = 0.75 \quad \quad y_B = 0.5 $$
at dew point
$$ \frac {1}{P_T } = \frac {Y_A }{P^0A} + \frac {Y_B}{P^0_B } \Rightarrow \frac {1}{ P_T} = \frac { 0.75}{400} + \frac { 0.25}{800} = \frac { 1.5 + 0.25}{800} $$
$$ \Rightarrow P_T = \frac {800}{1.75} = 457.14 mm Hg $$
Below dew point only vapor phase exists.
The reading of pressure gauge at which only vapour phase exists is
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0%
501
0%
457.14
0%
425
0%
525
Explanation
$$ X_A = 0.75 \quad X_B = 0.05 $$ above 500 mm Hg only liquid phase exists .
$$ P_{bubble} point = X_A P^0_A + X_B P^0_B $$
$$ P_{cqycqfyr fcUnq} = X_A P60_A + X_B P^0_B $$
$$ 0.75 +400 + 0.25 \times 800 = 500 mm $$
$$ y_A = 0.75 \quad \quad y_B = 0.5 $$
at dew point
$$ \frac {1}{P_T } = \frac {Y_A }{P^0A} + \frac {Y_B}{P^0_B } \Rightarrow \frac {1}{ P_T} = \frac { 0.75}{400} + \frac { 0.25}{800} = \frac { 1.5 + 0.25}{800} $$
$$ \Rightarrow P_T = \frac {800}{1.75} = 457.14 mm Hg $$
Below dew point only vapor phase exists.
The vapour pressure of two micible liquid A and B are 300 and 500 mm of Hf respectively. in a flask , 2 moles of A are mixed with moles of B. further to the mixture , 32 g of an ionic non -volatile solute MCI (partially ionised, mol mass = 70 u) were also added. thus, the final vapour pressure of solution was found to be 420 mm and Hg. Then, identify the correct statement(s). (Assume the liquid mixture of A and B to behave ideally).
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The numerical value of relative lowering in vapour pressure upon addition of solute MCI is 1/15
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Th solute MCI is 25 % ionised in the above question.
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The solute MCI is 23.33 % ionised in the above question
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Upon addition of excess $$ Pb(NO_3)_2 $$ the number of moles of $$ PbCl_2 $$ precipitated is 2/ 35 .
Explanation
$$ P_0= X_A P^0_A + X_B P^0_B = \frac {2}{8} \times 300 + \frac {6}{7} \times 500 = 450$$ mm of Hg
Now RLVP = $$ \frac {P_0 -P_s}{P_0} = \frac { 450 - 420}{450} = \frac {1}{15} $$
Also RLVP = $$ ( \frac {in}{In +N } ) $$
$$ \frac {1}{15} = \frac { \frac { 32i}{70} }{ \frac {32i}{70} + 8 } $$
$$ \therefore i - 1.25 = ( 1+ ( 2-1 ) \alpha ). $$.
So $$ \alpha $$ = 0.25 ( or 25 %)
$$ \therefore n_{CI^- } $$ produced = $$ \frac {25}{100} \times \frac {32}{70} = \frac {4}{35} $$
$$ \therefore n_{PbCl_2 } $$ precipitated = $$ \frac {1}{2} \times \frac {4}{35} = \frac {2}{35} $$
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