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Byju's Answer

Standard XII

Chemistry

Stoichiometric Calculations

The mole in 1...

Question

The mole in 16 g calcium is 0.4. What is the molar mass of Ca atoms?

4g/mole

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6.4 g/mole

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0.04 g/mole

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40 g/mole

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Solution

The correct option is C 40 g/mole
$n o . o f m o l e s = \frac{G i v e n m a s s}{M o l a r m a s s} A c c o r d i n g t o q u e s t i o n, 0.4 = \frac{16}{M o l a r m a s s} M o l a r M a s s = \frac{16}{0.4} = 40 g m / m o l e$

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Similar questions

40 g

of calcium was extracted from

56 g

of calcium oxide
(Atomic mass of

C a = 40, O = 16

)
i) What mass of oxygen is there in

56 g

of calcium oxide?
ii) How many moles of oxygen atoms are there in this?
iii) How many moles of calcium atoms are there in

40 g

of calcium?
iv) What mass of calcium will be obtained from

1000 g

of calcium oxide?

Q. Fill in the blanks using the given data:
The formula of calcium oxide is

C a O

. The atomic mass of

C a

40

, oxygen is

16

and carbon is

12

.
i)

1

mole of

C a

( ____g) and

1

mole of oxygen atom ( ___g) combine to form _____mole of

C a O

( ____g).
ii)

1

mole of

C a

( ___g) and

1

mole of

C

( ___g) and

3

moles of oxygen atom( ___g) combine to form

1 m o l e

C a C O_{3}

( ___g)

Q. How many moles of calcium carbonate

(C a C O_{3})

are present in 20 g of the substance?

(C a = 40, C = 12, O = 16)

Q. Statement 1: A

1.0

g sample of calcium citrate

C a_{3} (C_{6} H_{5} O_{7})_{2}

(molar mass

= 498

g/mol), contains more

C a

than a

1.0

g sample of calcium carbonate,

C a C O_{3}

(molar mass

= 100

g/mol)
Statement 2: There are more

C a

atoms in

1.0

mol of calcium carbonate than in

1.0

mol of calcium citrate.

Q. Molar ionic conductance of

{C a}^{+ 2}

x

S m^{2}

{m o l e}^{- 1}

. Equivalent conductance of calcium phosphate is ____

S m^{2} g

{e q}^{- 1}

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The mole in 16 g calcium is 0.4. What is the molar mass of Ca atoms?

The correct option is C 40 g/moleno.ofmoles=GivenmassMolarmassAccordingtoquestion,0.4=16MolarmassMolarMass=160.4=40 gm/mole

The correct option is C 40 g/mole
$n o . o f m o l e s = \frac{G i v e n m a s s}{M o l a r m a s s} A c c o r d i n g t o q u e s t i o n, 0.4 = \frac{16}{M o l a r m a s s} M o l a r M a s s = \frac{16}{0.4} = 40 g m / m o l e$