Question

What is the volume of a 4.0M solution that contains 374 grams of KNO3?

Answer 1

Answer:

$V_{solution}=0.925L=925mL$

Explanation:

Hello,

In this case, we should define what molarity is, being the ratio between the moles of the solute and the volume of the solution in litres:

$M=\frac{n_{solute}}{V_{solution}}$

Thus, the first step is to compute the moles of potassium nitrate, whose molar mass is 101.1 g/mol as shown below:

$n_{solute}=374g*\frac{1mol}{101.1g} =3.70molKNO_3$

Next, as we are asked for the volume of the solution we simply solve for it:

$V_{solution}=\frac{n_{solute}}{M}=\frac{3.70mol}{4.0mol/L} \\\\V_{solution}=0.925L=925mL$

Regards.

Answer 2

Answer:

0.93L

Explanation:

The following data were obtained from the question:

Molarity = 4M

Mass of KNO3 = 374g

Volume =..?

We'll begin by calculating the number of mole in 374g of KNO3. This is illustrated below:

Mass of KNO3 = 374g

Molar mass of KNO3 = 39 + 14 + (16x3) = 101g

Number of mole of KNO3 =..?

Number of mole = Mass/Molar Mass

Number of mole of KNO3 = 374/101 = 3.70 moles.

Now, we can obtain the volume of the solution as follow:

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 3.70/4

Volume = 0.93L

Therefore, the volume of the solution is 0.93L