The mass of 1- 10 22 molecules of Cu SO 4 -5 H 2 O is- Cu-63-5- S-32

The correct option is A 4.144 gmole= no. of molecules/6.022× 10^23 #160; #160; #160; #160; 1× 1022/6.022× 10^23 =0.0166 molemole of weight of ...

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Standard X

Chemistry

Stoichiometric Relation

The mass of ...

Question

The mass of $1 \times 10^{22}$ molecules of $C u {S O}_{4} .5 H_{2} O$ is: $(C u = 63.5, S = 32)$

4.144 g

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8.288 g

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2.648 g

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5.295 g

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Solution

The correct option is A 4.144 g
mole= $\frac{n o . o f m o l e c u l e s}{6.022 \times 10^{23}}$

$\frac{1 \times 1022}{6.022 \times 10^{23}}$ =0.0166 mole

mole of weight of $C U_{S} O_{4}$ , $5 H_{2} O =$ 0.0166\times 249.5 gm

=4.144 g

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Similar questions

Q. The molecular mass of

C u S O_{4} .5 H_{2} O

is:

(Atomic mass of

C u

=

63.5,

H

=

O

=

16,

S

=

32)

Q. The mass of

1 \times 10^{22}

molecules of blue vitriol

(C u S O_{4} \cdot x H_{2} O)

4.144

g. The value of

x

will be:

Q. The minimum quantity in grams of

H_{2} S

needed to precipitate

63.5 g

C u^{2 +}

ion will be:
(molar mass of

C u = 63.5 g/mol

)

C u^{2 +} + H_{2} S \to C u S + 2 H^{+}

Q. The minimum quantity in grams of

H_{2} S

needed to precipitate

63.5 g

C u^{2 +}

ion will be:
(molar mass of

C u = 63.5 g/mol

)

C u^{2 +} + H_{2} S \to C u S + 2 H^{+}

Q. If

2

millimoles of hydrated copper sulphate salt

(C u S O_{4} .5 H_{2} O)

is heated to dryness, then calculate the obtained mass of anhydrous copper sulphate salt.
(Given: molar mass of

C u = 63.5 g/mol

)

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The mass of 1×1022 molecules of CuSO4.5H2O is: (Cu=63.5,S=32)

The correct option is A 4.144 gmole=no.ofmolecules6.022×1023 1×10226.022×1023=0.0166 molemole of weight of CUSO4,5H2O=0.0166\times 249.5 gm =4.144 g

The mass of $1 \times 10^{22}$ molecules of $C u {S O}_{4} .5 H_{2} O$ is: $(C u = 63.5, S = 32)$

The correct option is A 4.144 g
mole= $\frac{n o . o f m o l e c u l e s}{6.022 \times 10^{23}}$

$\frac{1 \times 1022}{6.022 \times 10^{23}}$ =0.0166 mole

mole of weight of $C U_{S} O_{4}$ , $5 H_{2} O =$ 0.0166\times 249.5 gm

=4.144 g