How many grams of AgNO3 need to be dissolved in water to produce 50 mL

1.	How many grams of AgNO3 need to be dissolved in water to produce 50 mL of a 0.1 molar solution?Provide your answer in mg with 1 decimal place.
Answer» Given : Volume of Solution = 50 mL MOLARITY of solution = 0.1 M To find : Mass of silver NITRATE ( AgNO3 ) in MG Formula used : Molarity = Number of mole of solute ÷ Volume of Solution in L Number of mole = given mass ÷ molar mass Solution : Molar mass of silver nitrate = ( 1 × Atomic mass of Ag) + ( 1 × Atomic mass of N ) + ( 3 × Atomic mass of O ) ➝ Molar mass of silver nitrate = ( 1 × 108) + ( 1×14 ) + (3×16) ➝ Molar mass of silver nitrate = 108 + 14 + 48 ➝ Molar mass of silver nitrate = 170 g/mol ________________________________ Molarity = Number of mole of solute ÷ Volume of Solution in L ➝ Number of mole of silver nitrate = Molarity × volume of solution in L ➝ Number of mole of silver nitrate = 0.1 × (50/1000) ➝ Number of mole of silver nitrate = 0.005 moles ________________________________ Number of mole = REQUIRED mass ÷ molar mass ➝ Required mass = Number of mole of silver nitrate × molar mass of silver nitrate ➝ Required mass = 0.005 × 170 ➝ Required mass = 0.85g ➝ Required mass = 0.85g × 1000 mg/g ➝ Required mass = 850 mg ________________________________ ANSWER : 850 mg

How many grams of AgNO3 need to be dissolved in water to produce 50 mL of a 0.1 molar solution?Provide your answer in mg with 1 decimal place.

Answer»

Mass of silver NITRATE ( AgNO3 ) in MG

Molar mass of silver nitrate = ( 1 × Atomic mass of Ag) + ( 1 × Atomic mass of N ) + ( 3 × Atomic mass of O )

➝ Molar mass of silver nitrate = ( 1 × 108) + ( 1×14 ) + (3×16)

➝ Molar mass of silver nitrate = 108 + 14 + 48

➝ Molar mass of silver nitrate = 170 g/mol

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Molarity = Number of mole of solute ÷ Volume of Solution in L

➝ Number of mole of silver nitrate = Molarity × volume of solution in L

➝ Number of mole of silver nitrate = 0.1 × (50/1000)

➝ Number of mole of silver nitrate = 0.005 moles

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Number of mole = REQUIRED mass ÷ molar mass

➝ Required mass = Number of mole of silver nitrate × molar mass of silver nitrate

➝ Required mass = 0.005 × 170

➝ Required mass = 0.85g

➝ Required mass = 0.85g × 1000 mg/g

➝ Required mass = 850 mg

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