Describe the preparation of 2.000 L of 0.0500 M AgNO3 (169.87 g/mol) from the primary-standard-grade solid.

Transfer 1.6987 grams of AgNO₃ into a mixing container and add solvent up to but not to exceed 2.000 Liters total volume. Mix and apply as needed.

Explanation:

A simple formula for determining the amount of reagent needed for a specified solution from a manufacturer's 'solid' reagent** is ...

grams solute needed = Molarity x Volume of Solution in Liters x Molecular Weight of solute

Determination of mass of solute needed:

∴ grams AgNO₃ needed* = M·V·mol. wt = (0.0500M)(2.000L)(169.87g·mol⁻¹) = 1.6987 grams AgNO₃

*assumes 100% pure reagent.

Preparation:

Transfer 1.6987 grams of AgNO₃ into a mixing container and add solvent up to but not to exceed 2.000 Liters total volume. Mix and apply as needed.

NOTE => FYI => If a manufacturer's liquid form reagent is used, one must divide calculated grams by density of liquid supplied and gives volume in milliliters (ml) needed to prepare a specified solution.

Vol reagent needed for soln prep in milliliters = (Molarity·Volume in Liters·formula wt of reagent)/(density of stock liquid concentrate)

That is,  Vol(ml) = [(M)(V)(f.wt.)] / density of stock

Prep: Transfer needed volume (ml) of liquid stock into mixing vessel and add solvent up to but not to exceed total volume of solution needed.

NOTE => Caution when applying to strong acids. 1st add a small quantity of water such that water + volume of acid is less than total volume needed followed by strong acid, then dilute to the needed total volume. Such prevents solution from overheating because of high rate of ionization and flashing out of container.    

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

the answer is   c.) 1575 j.

explanation: