Question

A student wants to make a 0.150 M aqueous solution of silver (I) nitrate but only has 11.27 g of AgNO3. What volume of the 0.150 M solution can they make?

Answer 1

442.3 mL

Step-by-step explanation:

Remember that Molarity is a measure of concentration in Chemistry and it's defined as the number of moles of the substance divided by liters of the solution:

`M=(Moles of substance X)/(Volume of the solution)`

Then, you can express 11.27 g of AgNO3 as moles of AgNO3 using the molar mass of the compound:

`11.27 g AgNO_(3) *(1 mole AgNO_(3))/(169.87 g AgNO_(3)) = 0.06634 moles AgNO_(3)`

Then you can solve for the volume of the solution:

`Volume of the solution=(Moles of AgNO_(3))/(M) =(0.06634 mol AgNO_(3))/(0.150 M) =0.4423 L = 442.3 mL`

Hope it helps!