We add excess nacl solution (58.44 g/mol) to 72 ml of a solution of silver nitrate (agno3 169.88 g/mol), to form insoluble solid agcl. when it has been dried and weighed, the mass of agcl (143.32 g/mol) is found to be 2.61 grams. what is the molarity of the original agno3 solution? the formula weight of nano3 is 85.00 g/mol.

The molarity of the original AgNO3 solution is 0.2529 M

Explanation:

Step 1:  The balanced equation

AgNO3 + NaCL → AgCl + NaNO3

This means that for 1 mole of AgNO3 consumed, 1 mole of NaCL will be consumed, to produce 1 mole of AgCL and 1 mole of NaNO3

Step 2: Calculating moles of AgCl

Moles of AgCl = mass of AgCL / Molar mass of AgCL

moles of AgCl = 2.61 grams / 143.32g/mole

moles of AgCl = 0.01821 moles

Step 3: Calculating moles of AgNO3

Since there is consumed 1 mole of AgNO3 to produce 1 mole of AgCl, this means that if there is produced 0.01821 moles of AgCl, there is consumed also 0.01821 moles of AgNO3

Step 4: Calculating molarity of AgNO3

Molarity of AgNO3 = moles of AgNO3 / volume of AgNO3

Molarity of AgNO3 = 0.01821 moles/  0.072 L = 0.2529 M = 0.2529 mol/L

The molarity of the original AgNO3 solution is 0.2529 M

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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