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Standard XII

Chemistry

Types of Solution

Number of mil...

Question

Number of millilitres of a $1.6 %$ $B a C l_{2} (w / v)$ solution which is required to precipitate sulphur as $B a S O_{4}$ in a $0.60$ g sample that contains $12 %$ sulphur is :

58.50

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14.63

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29.25

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21.00

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Solution

The correct option is A $29.25$ mL
$S ⟶ B a S O_{4} ⟶ B a C l_{2}$
$32$ g $208$ g
$0.60 \times 0.12 = 0.072$ g
$B a C l_{2}$ , required by $0.072$ g sulphur to be precipitated as,
$B a S O_{4} = \frac{208}{32} \times 0.072$ $= 0.468 g m$
$T h u s, v o l u m e o f B a C l_{2} s o l u t i o n = \frac{0.468 \times 100}{1.6}$ $= 29.25 m L$

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Similar questions

Q. Number of mililiters of a

1.6 %

B a C l_{2} (w / v)

solution, which is required to precipitate the sulphur as

B a S O_{4}

in a

0.60

g sample that contains

12 %

S

, is:

Q. Calculate the number of millilitres of

N H_{3} (a q)

solution

(d = 0.986

g/ml

)

containing

2.5 %

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N H_{3}

, which will be required to precipitate iron as

F e (O H)_{3}

in a

0.8 g

sample that contains

50 % F e_{2} O_{3}

Q. Calculate the number of millilitres of

N H_{3} (a q)

solution (d =

0.986

g/mL) containing

2.5 %

by weight

N H_{3}

, which will be required to precipitate iron as

F e (O H)_{3}

in a

0.8

g sample that contains

50 % F e_{2} O_{3}

(as nearest integer).

1.249

g of a sample of pure

B a C O_{3}

and impure

C a C O_{3}

containing some

C a O

was treated with dilute

H C l

and it evolved

168

mL of

C O_{2}

at NTP. From this solution,

B a C r O_{4}

was precipitated, filtered and washed. The dried precipitate was dissolved in dilute sulphuric acid and diluted to

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10

mL of this solution when treated with

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solution liberated iodine, which required exactly

20

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C a O

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1.6

g of pyrolusite ore was treated with

50

mL of

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oxalic acid and some sulphuric acid. The oxalic acid left in excess was raised to

250

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Number of millilitres of a 1.6% BaCl2(w/v) solution which is required to precipitate sulphur as BaSO4 in a 0.60 g sample that contains 12% sulphur is :

The correct option is A 29.25 mLS⟶BaSO4⟶BaCl232 g 208 g0.60×0.12=0.072 g BaCl2, required by 0.072 g sulphur to be precipitated as,BaSO4=20832×0.072=0.468gmThus,volumeofBaCl2solution=0.468×1001.6=29.25mL

Number of millilitres of a $1.6 %$ $B a C l_{2} (w / v)$ solution which is required to precipitate sulphur as $B a S O_{4}$ in a $0.60$ g sample that contains $12 %$ sulphur is :