Question

. Calculate the number of grams of silver nitrate required to prepare 2.00L of 0.50M AgNO3.

Answer 1

Ans: massAgNO3=formulamassAgNO3*Molarity*volume

Answer 2

MOLARITY

• Molarity is the measure of moles of solutes present in a solution.
• It is expressed in Moles per unit litres of a solution.

• $Molarity = \frac{Number of moles}{Volume of solution}$

• Number of moles $(n) = \frac{Given weight (w)}{Molecular weight}$

Given -

Molarity (M) = 0.5

Volume of solution = 2L

Molecular weight = 169.87 g/mol

Putting the values in the formulae,   $0.5 = \frac{n}{2}$

n = 1 mole

Given weight (w) = n × Molecular weight

w = 1 × 169.87

w = 169.87 g

169.87 grams of Silver nitrate is required to prepare 2L of 0.5M  of silver nitrate solution.