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viva [34]
3 years ago
7

How many grams are in 1 mole of Pb3(PO4)4?

Chemistry
2 answers:
Snezhnost [94]3 years ago
8 0

Answer:

Explanation:

Molar mass of Pb3(PO4)4 is 380+621.6 =1001.6 g/mole

 Molar comcentration = Reacting mass / Molar Mass

                       1 = Mr/1001.6

                       Mr = 1001.6g

kvasek [131]3 years ago
6 0

Answer:

In 1 mol of Pb₃(PO₄)₄ occupies 1001.48 grams

Explanation:

This compound is the lead (IV) phosphate.

Grams that occupy 1 mole, means the molar mass of the compound

Pb = 207.2  .3 = 621.6 g/m

P = 30.97 .4 = 123.88 g/m

O = (16 .  4) . 4 = 256 g/m

621.6 g/m + 123.88 g/m + 256 g/m = 1001.48 g/m

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Formula to calculate osmotic pressure is as follows.

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  \frac{\text{moles}}{\text{volume (in L)}} = 0.0285

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}

Now, let us assume that mass of C_{12}H_{23}O_{5}N = x grams

And, mass of C_{12}H{22}O_{11} = (1.00 - x)

So, moles of C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}

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Therefore, we can conclude that amount of C_{12}H_{23}O_{5}N present is 0.346 g  and amount of C_{12}H_{22}O_{11} present is (1 - 0.346) g = 0.654 g.

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