Solutions Test 39

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Solutions Test 39

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Question 1
1 / -0

When the air temperature is below freezing, the saturation vapor pressure over water is _______.

A
Equal to zero

B
Less than the saturation vapor pressure over ice

C
Greater than the saturation vapor pressure over ice

D
Equal to the saturation vapor pressure over ice
Question 2
1 / -0

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

A
$$65\%$$

B
$$69\%$$

C
$$76\%$$

D
$$98\%$$

Solution
Question 3
1 / -0

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.

A
$$500, 0.4, 0.6$$

B
$$500, 0.5, 0.5$$

C
$$450, 0.4, 0.6$$

D
$$450, 0.5, 0.5$$

Solution

$$P_T = X_AP_{A^o} + X_BP_{B^o}$$
$$= 0.5\times 400+0.5\times 600$$
$$=500nm$$ of $$Hg$$
$$\dfrac{1}{P_T} = \dfrac{y_A}{P^o_A} + \dfrac{1-y_B}{P^o_B}$$
$$y_A = 0.6$$ & $$y_B = 1 -0.6 = 0.4$$
Question 4
1 / -0

How many types of solutions are possible by the combination of any two of three states of matter (solid, liquid, and gas)

A
$$3$$

B
$$6$$

C
$$9$$

D
$$18$$

Solution
Question 5
1 / -0

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

A
$$i = 1 - x$$

B
$$i = 1 + \dfrac{x}{n}$$

C
$$i = \dfrac{i - x + \dfrac{x}{n}}{1}$$

D
$$i = 1$$

Solution
Question 6
1 / -0

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 hydrogen by wt.in vitamin $$C$$ ?

A
$$4.55\%$$

B
$$41\%$$

C
$$20.5\%$$

D
$$9.11\%$$

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 $$=1- (0.41+0.0455)=0.544\ g$$
$$\% C=\dfrac{0.41}{18}\times 100=41\%$$
$$\%H =\dfrac{0.0455}{1}\times 100=4.55\%$$ Option $$(a)$$
Question 7
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Directions For Questions

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

...view full instructions

How much water is evaporated per hour?

A
$$500\ kg$$

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 $$=20,000-500=19,500$$ Option $$(c)$$
Question 8
1 / -0

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 $$mol^{-1}$$)

A
$$3.72$$

B
$$1.83$$

C
$$6.8$$

D
$$1.86$$

Solution

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.
Question 9
1 / -0

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? $$(Sn=119)$$

A
$$33.33\%$$

B
$$26.27\%$$

C
$$29.38\%$$

D
$$52.54\%$$

Solution

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 $$Sn=26.27\%$$
Question 10
1 / -0

Directions For Questions

When the Bayer's process is used for recovering aluminium from siliceous ores, some aluminium is always lost because of the formation of an unworkable mud having the following average formula: $$3Na_{2}O.3Al_{2}O_{3}.5SiO_{2}.5H_{2}O$$. Since aluminium and sodium ions are always in excess in the solution from which this precipitate is formed, the precipitate of the silicon in the mud is complete. A certain ore contains $$13\%$$ (by weight) Kaolin, $$Al_{2}O_{3}.2SiO_{2}.2H_{2}O$$ and $$87\%$$ gibbsite, $$Al_{2}O_{3}.3H_{2}O$$. $$(Al=27, Si=28)$$

...view full instructions

What is the percentage of silica present in the ore, by weight?

A
$$2.82$$

B
$$3.02$$

C
$$0.465$$

D
$$6.05$$

Solution

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 $$(d)$$

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Solutions Test 39

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Solutions Test 39

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SHARING IS CARING

Question 1

Equal to zero

Less than the saturation vapor pressure over ice

Greater than the saturation vapor pressure over ice

Equal to the saturation vapor pressure over ice

Question 2

$$65\%$$

$$69\%$$

$$76\%$$

$$98\%$$

Solution

Question 3

$$500, 0.4, 0.6$$

$$500, 0.5, 0.5$$

$$450, 0.4, 0.6$$

$$450, 0.5, 0.5$$

Solution

Question 4

$$3$$

$$6$$

$$9$$

$$18$$

Solution

Question 5

$$i = 1 - x$$

$$i = 1 + \dfrac{x}{n}$$

$$i = \dfrac{i - x + \dfrac{x}{n}}{1}$$

$$i = 1$$

Solution

Question 6

$$4.55\%$$

$$41\%$$

$$20.5\%$$

$$9.11\%$$

Solution

Question 7

$$500\ kg$$

$$500\ kg$$

$$19,500\ kg$$

$$20, 000\ kg$$

Solution

Question 8

$$3.72$$

$$1.83$$

$$6.8$$

$$1.86$$

Solution

Question 9

$$33.33\%$$

$$26.27\%$$

$$29.38\%$$

$$52.54\%$$

Solution

Question 10

$$2.82$$

$$3.02$$

$$0.465$$

$$6.05$$

Solution

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