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Willchemistry
1 mole --------- 6.02 x 10²³ molecules
? mole --------- 4.00 x 10²³ molecules

moles =  (4.00 x 10²³) x 1 / 6.02 x 10²³

moles = ( 4.00 x 10²³) / ( 6.02 x 10²³ )

= 0.6644 moles

Molar mass H₃PO₄ = 98.0 g/mol

mass = 0.6644 x 98.0 = 65.1112 g

hope this helps!

Answer : The approximate mass of [tex]H_3PO_4[/tex] is 65 grams.

Explanation :

First we have to calculate the moles of [tex]H_3PO_4[/tex].

As, [tex]6.022\times 10^{23}[/tex] molecules of [tex]H_3PO_4[/tex] present in 1 mole of [tex]H_3PO_4[/tex]

So, [tex]4.00\times 10^{23}[/tex] molecules of [tex]H_3PO_4[/tex] present in [tex]\frac{4.00\times 10^{23}}{6.200\times 10^{23}}=0.664[/tex] mole of [tex]H_3PO_4[/tex]

Now we have to calculate the mass of [tex]H_3PO_4[/tex].

[tex]\text{Mass of }H_3PO_4=\text{Moles of }H_3PO_4\times \text{Molar mass of }H_3PO_4[/tex]

Molar mass of [tex]H_3PO_4[/tex] = 98 g/mole

[tex]\text{Mass of }H_3PO_4=0.664mole\times 98g/mole=65.072g\approx 65g[/tex]

Therefore, the approximate mass of [tex]H_3PO_4[/tex] is 65 grams.

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