An experiment requires 0.38 M NH3(aq). The stockroom manager estimates that 15 L of the base is needed. What volume of 15 M NH3(aq) will be used to prepare this amount of 0.38 M base? Answer in units of mL.

480 mL

Explanation:

15L  of  0.38M of NH3(aq.) means that we require 15x0.38 moles of NH3.

this is to be provided by 15M x V.  

Where V is volume in Liters.

These are equal  

⇒15M x V = 15x 0.48  

Hence, Volume required V = 0.48L =480mL

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.

let that person have a turn to speak : )