Question

You make 1.000L of an aqueous solution that contains 35.0 g of sucrose (C12 H22 O11) a.) what is the molarity of sucrose in this solution? 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

a)

Molar mass of C12H22O11,

MM = 12*MM(C) + 22*MM(H) + 11*MM(O)

= 12*12.01 + 22*1.008 + 11*16.0

= 342.296 g/mol

mass(C12H22O11)= 35 g

use:

number of mol of C12H22O11,

n = mass of C12H22O11/molar mass of C12H22O11

=(35 g)/(3.423*10^2 g/mol)

= 0.1023 mol

volume , V = 1 L

use:

Molarity,

M = number of mol / volume in L

= 0.1023/1

= 0.1023 M

Answer: 0.102 M

b)

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 0.102 M

M2 = 0.051 M

V1 = 1 L

use:

M1*V1 = M2*V2

V2 = (M1 * V1) / M2

V2 = (0.102*1)/0.051

V2 = 2 L

This is final volume

Volume of water added = V2 - V1

= 2 L - 1 L

= 1 L

Answer: 1 L