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Standard XII

Chemistry

Density

A metal has a...

Question

A metal has a fcc lattice. The edge length of the cell is 404 pm. The density of the metal is $2.72 g c m^{- 3}$ . The molar mass of the metal is
( $N_{A}$ Avogadro's constant = $6.02 \times 10^{23} m o l^{- 1}$ )

$27 g m o l^{- 1}$

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$20 g m o l^{- 1}$

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$40 g m o l^{- 1}$

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$30 g m o l^{- 1}$

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Solution

The correct option is A
$27 g m o l^{- 1}$

M = $\frac{d . a^{3} . N_{A}}{Z}$ = $\frac{2.72 \times [4.4 \times 10^{- 10}]^{3} \times 6.02 \times 10^{23}}{4} = 26.9 g m o l^{- 1} \approx 27 g m o l^{- 1}$

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Similar questions

A metal has a fcc lattice . The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm^-3. The molar mass of the metal is:

(N_A Avogadro's constant =6.02×10²³ mol^-1)

A metal has a fcc lattice. The edge length of the cell is 404 pm. The density of the metal is $2.72 g c m^{- 3}$ . The molar mass of the metal is
( $N_{A}$ Avogadro's constant = $6.02 \times 10^{23} m o l^{- 1}$ )

Q. Metal has an fcc lattice. The edge length of the unit cell is

404

pm. The density of the metal is

2.72

g cm

^{- 3}

. The molar mass (in g mol

^{- 1}

) of the metal is:
[Avogadro's constant

= N_{A} = 6.02 \times 10^{23}

mol

^{- 1}

]

Q. A metal has afcc lattice. The edge length of the unit cell is

404 pm

. The density of the metal is

2.72 {g cm}^{- 3}

.
The molar mass of the metal is (

N_{A}

Avogadro's constant =

6.02 \times 10^{23} {mol}^{- 1}

)

Q. A metal has fcc lattice. The edge length of the unit cell is 404pm. The density of metal is 2.72 g/ cm. The molar mass of the metal is (N=6.02\times 10 mol)

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A metal has a fcc lattice. The edge length of the cell is 404 pm. The density of the metal is 2.72 g cm−3 . The molar mass of the metal is (NA Avogadro's constant = 6.02×1023mol−1)

The correct option is A 27 g mol−1 M =d.a3.NAZ =2.72×[4.4×10−10]3×6.02×10234=26.9gmol−1≈27gmol−1

A metal has a fcc lattice. The edge length of the cell is 404 pm. The density of the metal is $2.72 g c m^{- 3}$ . The molar mass of the metal is
( $N_{A}$ Avogadro's constant = $6.02 \times 10^{23} m o l^{- 1}$ )

The correct option is A
$27 g m o l^{- 1}$

M = $\frac{d . a^{3} . N_{A}}{Z}$ = $\frac{2.72 \times [4.4 \times 10^{- 10}]^{3} \times 6.02 \times 10^{23}}{4} = 26.9 g m o l^{- 1} \approx 27 g m o l^{- 1}$