How many grams are there in 7.4 x 10^23 molecules of agno3?

Data we are collecting from the equation as:

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

As we know avogandro'S no. 6.022*1023 which is number of molecules.

So when we calculate any no. Of mole of any atom we know that 1mole of any atom is equal to avogandro'S no and grams can be calculate from moles equation.

So back to 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

Weight of AgNO3 is 163.868.

Now multiplying both values moles*molar weight of AgNO3

1.229*1023 * 763.868 = 208.767grams of AgNo3

pop pop pop on a opp