Question

Molarity problem?

You make 1.000 L of an aqueous solution that contains 35.0 g of sucrose (C12H22O11).

A) What is the molarity of sucrose in this solution? (I got 0.102 M for Part A) I just can't figure out part B.

B) How many liters of water would you have to add to this solution to reduce the molarity you calculated in Part A by a factor of two?

Answer 1

By definition....`"Molarity"="Moles of solute"/"Volume of solution"`

Explanation:

And if the solute is sucrose we got....

`((35.0*g)/(342.3*g*mol^-1))/(1.0*L)=0.102*mol*L^-1` with respect to sucrose.....

And given the defining quotient...we may repeat it again....

`((35.0*g)/(342.3*g*mol^-1))/(2.0*L)=0.051*mol*L^-1`

And another way we could this is to use the old relationship:

`C_1V_1=C_2V_2`, where `V_i="volume"`, and `C_i="concentration"`

`C_2=(C_1V_1("moles of solute"))/(V_2("the new volume"))=0.0510*mol*L^-1` as required.....

Capisce?