Question

Calculate the molarity of a solution in which 212.5g of NaNO3 are contained in 3.0 liters of solution.

Answer 1

Molarity's formula is: moles solute/liters solution.

In this question, we are given grams of solute so we have to convert that to moles.

NaNO3 has a molar mass of 84.9947g/mol. Here we begin with the given 212.5g and multiplying it by 1mol/84.9947g because the grams have to cancel out to give moles.

$(212.5 grams NaNO3)/(\frac{1 mole NaNO3}{84.9947 grams NaNO3} ) =2.5 moles NaNO3$

Now that we have moles of solute, we just plug it into the formula.

2.5 mols NaNO3/3L solution=0.8333 M of solution

Answer 2

The molarity of the solution containing 212.5 g of NaNO₃ in 3 L of solution is 0.83 M

What is molarity?

This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

How to determine the mole of NaNO₃

• Mass of NaNO₃ = 212.5 g
• Molar mass of NaNO₃ = 23 + 14 + (3×16) = 85 g/mol
• Mole of NaNO₃ =?

Mole = mass / molar mass

Mole of NaNO₃ = 212.5 / 85

Mole of NaNO₃ = 2.5 mole

How to determine the molarity

• Mole of NaNO₃ = 2.5 mole

• Volume = 3 L

• Molarity =?

Molarity = mole / Volume

Molarity = 2.5 / 3

Molarity = 0.83 M

Learn more about molarity:

https://brainly.com/question/15370276

#SPJ2