# How many mLs of 18.0 M H2SO4 must be added to 100mL H2O to give soln of 5.00M H2SO4?

I don't know how to set this up or solve, need some help.

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Seems like a trick question!!!!!

Volumes ARE NOT ADDITIVE, but assuming that they were, how many mLs of 18.0M H2SO4 must be ADDED to 100mL of H2O to give a solution of 5.00M H2SO4?

Use the following relationship:

M1V1=M2V2

Where:

M1=18M

V1=?

M2=5.0M

V2=100mL+V1

(18M)*(V1)=(5M)*(0.1L +V1)

V1=[(5M)*(0.1L +V1)]/18M

V1=[500M*L+5M*V1]/18M

V1=27.78mL+0.2778V1

Subtract 0.2778V1 from both sides of the equation to isolate V1's and mL's on opposite sides of the equation.

V1-0.2778V1=27.78mL+0.2778V1-0.2778V1

0.7222V1=27.78mL

0.7222/0.7222V1=27.78mL/0.7222

V1=38.46mL=38.5mL

How many moles does 38.5mL of 18M H2SO4 contain?

38.5mL=38.5 x 10^-3L

Moles=Molarity*volume(L)

moles=18M*38.5 x 10^-3L=0.6923 moles of H2SO4

M=moles/volume(L)=0.6923 moles/(0.0385L+0.100L)=4.998M=5.00M

However, if you meant how many mLs of 18M H2SO4 do you need to make a 100mL solution of 5.00M H2SO4 solution, then use the following relationship:

M1V1=M2V2

Where:

M1=18M

V1=?

M2=5.0M

V2=100mL

(18M)*(V1)=(5M)*(100mL)

Solve for V1

V1=(5M/18M)*(100mL)

V1=27.77mL=27.8mL

Add 50mL of H2O followed by 27.8mL of 18M acid. Fill the rest of volumetric flask with H2O until the bottom of the meniscus touches the line and shake throughly.

Best.

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• The response of SO3 with H2O produces a mist of H2SO4 which corrodes the containers as it is rather reactive . The response is violent and can exit of control.The mist is hard to isolate.The 2nd replacement procedure which is used more almost always is cheaper and more efficient.Additionally the acid can be diluted to the preferred awareness within the process.

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• M1V1 = M2V2

M1 = 5.00 M

V1 = 0.100 L

M2 = 18.0 M

solve for V2

V2 = M1V1/M2 = 5.00 * 0.100 / 18.0 = 0.0278 L or 27.8 mL

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