Question

(a) Find the number of Na ions in 2.92 g of Na2co3​

Answer 1

Given ;

• W(Na2CO3) = 2.92g

M(Na2CO3) =

• 2(Na) + 1(C) + 3(O)
• 2(23) + 1(12) + 3(16)
• 46 + 12 + 48
• 46 + 60
• 106 g/mol.

Thus moles of Na2CO3;

• n = W/M
• n = 2.92/106
• n ≈ 0.02 moles.

In 1 mole of Na2CO3 => 2 Na ions

Thus in 0.02 moles => x ions

By unitary method;

• x = 0.02 × 2
• x = 0.04 moles.

Thus number of ions = n×NA

• N = n×NA
• N = 0.04 × 6.022 × 10²³
• N = 0.24088 × 10²³
• N = 24.088 × 10²¹ ions

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