Question

Question 3 (1 point) Solution Composition: Molarity and Volume How many milliliters of (2.45x10^-1) M RbHCO3 solution contains (2.069x10^1) g of the solute? (Example answer: 12.3 or 1.23e-3) Note: Your answer is assumed to be reduced to the highest power possible. Your Answer:

Answer 1

Molar mass of RbHCO3,

MM = 1*MM(Rb) + 1*MM(H) + 1*MM(C) + 3*MM(O)

= 1*85.47 + 1*1.008 + 1*12.01 + 3*16.0

= 146.488 g/mol

mass(RbHCO3)= 20.69 g

use:

number of mol of RbHCO3,

n = mass of RbHCO3/molar mass of RbHCO3

=(20.69 g)/(1.465*10^2 g/mol)

= 0.1412 mol

use:

M = number of mol / volume in L

0.245 = 0.1412/ volume in L

volume = 0.576491 L

volume = 5.765*10^2 mL

Answer: 5.76e2 mL