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Question

How much sample of $B a C l_{2} . 2 H_{2} O$ will you take, if you need 0.75 mol of $B a^{2 +}$ ions?
$(B a = 137, C l = 35.5, H = 1, O = 16)$

183 g

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180 g

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283 g

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383 g

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Solution

The correct option is A 183 g
$B a C l_{2} . 2 H_{2} O$
Molar mass $= 137 + 35.5 \times 2 + 2 \times 18$
$= 244$
moles $= \frac{g r a m w e i g h t}{a t o m i c m a s s}$
$0.75 = \frac{g r a m w e i g h t}{244}$
{moles of $B a^{+ 2} =$ moles of $B a C l_{2} . 2 H_{2} O$ }
gram weight = $183 g$

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Similar questions

Q. How much sample of

B a C l_{2} .2 H_{2} O

should be taken to get 0.75 mole of

B a^{2 +}

ions?

(B a = 137; C l = 35.5; H = 1; O = 16)

Q. The amount of

B a S O_{4}

formed upon mixing

100

ml of

20.8 %

B a C l_{2}

solution with

50

ml of

9.8 %

H_{2} S O_{4}

solution will be:

(Given that molecular weight of

B a = 137, C l = 35.5, S = 32, H = 1

and

O = 16

g/mol)

Q. 100 mL of 20.8%

B a C l_{2}

solution and 50 mL of 9.8%

H_{2} S O_{4}

solution will form

B a S O_{4}

(B a = 137, C l = 35.5, S = 32, H = 1, O = 16)

B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l

Q. The amount of

B a S O_{4}

formed upon mixing

100

mL of

20.8 %

B a C l_{2}

solution with

50

mL of

9.8 %

H_{2} S O_{4}

solution will be:
[Ba

= 137

, Cl

= 35.5

, S

= 32

, H

= 1

and O

= 16

Q. The number of moles of water in 976 g

{B a C l}_{2} 2 H_{2} O

is:

(atomic masses of

H, O, C l,

and

B a

are 1, 16, 35, 5, and 137 respectively)

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How much sample of BaCl2.2H2O will you take, if you need 0.75 mol of Ba2+ ions?(Ba=137, Cl=35.5, H=1, O=16)

The correct option is A 183 gBaCl2.2H2OMolar mass =137+35.5×2+2×18=244moles =gramweightatomicmass0.75=gramweight244{moles of Ba+2= moles of BaCl2.2H2O}gram weight = 183g

How much sample of $B a C l_{2} . 2 H_{2} O$ will you take, if you need 0.75 mol of $B a^{2 +}$ ions?
$(B a = 137, C l = 35.5, H = 1, O = 16)$

The correct option is A 183 g
$B a C l_{2} . 2 H_{2} O$
Molar mass $= 137 + 35.5 \times 2 + 2 \times 18$
$= 244$
moles $= \frac{g r a m w e i g h t}{a t o m i c m a s s}$
$0.75 = \frac{g r a m w e i g h t}{244}$
{moles of $B a^{+ 2} =$ moles of $B a C l_{2} . 2 H_{2} O$ }
gram weight = $183 g$