Question

How many are molecules ( or formula) in each sample?

55.93 kg NaHCO3

4.59•10^5 g Na3PO4​

Answer 1

• 
  `4.010 * 10^(25) \text { molecules of } \mathrm{NaHCO}_(3) </strong>\text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_(3)`
• 
  `16.86 * 10^(26) \text { molecules of } \mathrm{Na}_(3) \mathrm{PO}_(4) </strong>\text { present in } 459 \mathrm{kg}\left(4.59 * 10^(5) \mathrm{gm}\right) \text { of } \mathrm{Na}_(3) \mathrm{PO}_(4)`

Step-by-step explanation:

Number of molecules for
`55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_(3)`

`\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_(3):`

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

`\text { Number of molecules of } \mathrm{NaHCO}_(3) \text { in } 55.93 \text { kg are as follows: }`

`55.93 *\left(10^(3) \mathrm{gm}\right) * \frac{1 \mathrm{mol} \mathrm{NaHCO}_(3)}{84.00 \mathrm{gm} \mathrm{NaHCO}_(3)} *\left(6.022 * 10^(23) \mathrm{molecules} \text { i.e Avogadro number }\right)`

`=4.010 * 10^(26) \text { molecules of } \mathrm{NaHCO}_(3) \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_(3)`

Number of molecules for for
`\left(4.59 * 10^(5) \mathrm{gm}\right) \text { of } \mathrm{Na}_(3) \mathrm{PO}_(4)`

`\text { Firstly molar mass is calculated of } \mathrm{Na}_(3) \mathrm{PO}_(4)`

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

`459 *\left(10^(3) \mathrm{gm}\right) * \frac{1 \mathrm{mol} N a_(3) P O_(4)}{163.94 \mathrm{gm} N a_(3) P O_(4)} *\left(6.022 * 10^(23) \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.`

`=16.86 * 10^(26) \text { molecules of } \mathrm{Na}_(3) \mathrm{PO}_(4) \text { present in } 459 \mathrm{kg}\left(4.59 * 10^(5) \mathrm{gm}\right) \text { of } \mathrm{Na}_(3) \mathrm{PO}_(4)`