Question

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 1

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

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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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ANSWER : 850 mg