Question

Calculate the number of atoms of the constituent elements in 53 g of Na2CO3.​

Answer 1

Answer:

53 g of Na2CO3 Contain = 53106=12 mol.

No of atoms of Na = 12×2×6.022×1023= 6.022×1023.

No of atoms of C = 12×6.022×1023.

No of atoms of O = 12×3×6.022×1023= 9.033×1023.

please give my ans in brain mark list

Answer 2

Explanation:

${\huge{\sf{\fbox{\orange{Question}}}}}$

Calculate the number of atoms of the constituent elements in 53 g of Na2CO3.

$\\ {\huge{\sf{\fbox{\pink{Answer}}}}}$

Molecular mass of Na2CO3,

${\bold{\sf{➠ \: ( 2 × 23 ) \: + \: 12 \: + \: (3 × 16)}}}$

${\bold{\sf{➠ \: 46 \: + \: 12 \: + \: 48 = \: 106}}}$

Given mass of Na2CO3 = 53 g

106 g of Na2CO3 Contain = 1 mol

${\bold{\sf{53 g \: of \: Na2CO3 \: Contain \: = \: \frac{53}{106} \: = \: \frac{1}{2} \: mol}}}$

No of atoms of Na,

${\bold{\sf{➠ \: \frac{1}{2} \: × \: 2 \: × \: 6.022 \: × \: 10²³}}}$

${\bold{\sf{➠ \: 6.022 \: × \: 10²³}}}$

$\\ {\bold{\sf{No. \: of \: atoms \: of \: C \: = \: \frac{1}{2} \: × \: 6.022 \: × \: 10²³}}}$

${\bold{\sf{➠ \: 3.011 \: × \: 10²³}}}$

No of atoms of O,

${\bold{\sf{➠ \: \frac{1}{2} \: × \: 3 \: × \: 6.022 \: × \: 10²³}}}$

${\bold{\sf{➠ \: 9.033 \: × \: 10²³}}}$

${\large{\sf{\fbox{\red{Hence, the number of atoms in Na2CO3 is 9.033 × 10²³}}}}}</p><p>$