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How many grams are there in 7.4 x 1023 formula units of AgNO3?

User Urbanmojo
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1 Answer

21 votes
21 votes

Data we are collecting from the equation as:

7.4x10^23 = silver nitrate molecules as given in the question.

As we know Avogadro's no. 6.022*1023 which is a number of molecules.

So when we calculate any no. Of the mole of any atom we know that 1mole of any atom is equal to Avogadro's no and grams can be calculated from the mole's equation.

So back to the point,

Dividing molecules of AgNO3 to no. of molecules per mol of AgNO3

7.4 / 6.02*1023

= 1.229*1023per mole of AgNO3.

OK we get the moles of AgNO3

Now checking the gram of AgNO3

The weight of AgNO3 is 163.868.

Now multiplying both values moles*molar weight of AgNO3

1.229*1023 * 763.868 = 208.767grams of AgNo3

User Pabloasc
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