Question

Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L

Answer 1

Answer:

The answer is 1.15m.

Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H2SO4 and the mass of the solvent, which I presume is water.

We can find the number of H2SO4 moles by using its molarity

C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288

Since water has a density of 1.00kgL, the mass of solvent is

m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg

Therefore, molality is

m=nmass.solvent=0.288moles0.250kg=1.15m

Answer 2

Answer:

The molarity of 48.0 mL of 6.00 M H₂SO₄ diluted to 0.250 L is 1.152 M

Explanation:

Dilution is the procedure used to prepare a less concentrated solution from a more concentrated solution. It consists of adding solvent to an existing solution. Then the amount of solute does not vary, but the volume of the solvent does: when more solvent is added, the concentration of the solute decreases, as the volume of the solution increases.

When working with dilutions, the following can be applied, taking into account that the mass of the solute does not change and neither does its moles:

moles of solute before dilution = moles of solute after dilution

Mi*Vi=Mf*Vf

where

• Mi: initial molarity of the solution.
• Vi: initial volume of the solution.
• Mf: final molarity of the solution.
• Vf: final volume of the solution.

In this case, the following values ​​are possessed:

• Mi= 6 M
• Vi= 48 mL=0.048 L (1 L=1000 mL)
• Mf= ?
• Vf= 0.250 L

Replacing:

6 M*0.048 L=Mf* 0.250 L

Resolving:

$Mf=\frac{6M*0.048L}{0.250 L}$

Mf=1.152 M

The molarity of 48.0 mL of 6.00 M H₂SO₄ diluted to 0.250 L is 1.152 M