A chemist studying the properties of photographic emulsions needed to prepare 500 mL of 0.155 M AgNO3(aq). Wha?

This is a pretty generic dilutions problem. You have the initial molarity of the solution, and the volume that you want. to find the mass of the AgNO3(aq) you use the formula (Molarity)(Volume in Liters)=moles. So: (.155)(.500)=.0775 moles AgNO3. Then, you just change that to grams: .0775 mol X (170 g AgNO3/mol)=13.2 g btw, the equation makes sense because Molarity is (moles substance/Liters solution), so if you multiply Molarity times the volume of the solution you will get the number of moles of the substance you are using. Hope this helped!

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Use the information to find the number of moles of AgNO3 in the solution moles = Conc x Vol moles = 0.155 mol/L x 0.500 L = 0.0775 moles Now convert that many moles of AgNO3 into a mass mass = moles x Molar mass mass = 0.0775 moles x 169.88 g/mol = 13.17 g

This is a pretty generic dilutions problem. You have the initial molarity of the solution, and the volume that you want. to find the mass of the AgNO3(aq) you use the formula (Molarity)(Volume in Liters)=moles. So: (.155)(.500)=.0775 moles AgNO3. Then, you just change that to grams: .0775 mol X (170 g AgNO3/mol)=13.2 g btw, the equation makes sense because Molarity is (moles substance/Liters solution), so if you multiply Molarity times the volume of the solution you will get the number of moles of the substance you are using. Hope this helped!

Use the information to find the number of moles of AgNO3 in the solution moles = Conc x Vol moles = 0.155 mol/L x 0.500 L = 0.0775 moles Now convert that many moles of AgNO3 into a mass mass = moles x Molar mass mass = 0.0775 moles x 169.88 g/mol = 13.17 g

moles of AgNO3 = .168 x 500 /1000 = 0.084 moles Molar mass of AgNO3 is 169.8732 g/mol so mass of AgNO3 = 14.269g of the silver salt