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Solid Solute and Molarity

The moles that are required can be calculated from the volume (liters) and molarity (moles per liter)  [Pg.70]

The liters in Equation (4.10) are the liters of solution that are desired the moles per liter is the molarity desired and the grams per mole is the formula weight of the solute. Thus, we have [Pg.70]

How would you prepare 500.0 mL of a 0.20 M solution of NaOH from pure, solid NaOH  [Pg.71]

The analyst would weigh 4.0 g of NaOH, place it in a container with a 500-mL calibration line, add water to dissolve the solid, and then dilute to volume and shake, stir, or swirl. [Pg.71]

If the solute is a liquid (somewhat rare), the weight calculated from Equation (4.11) can, but rather inconveniently, be measured on a balance. However, this weight can also be converted to milliliters by using the density of the liquid. In this way, the volume of the liquid can be measured, rather than its weight, and it can be pipetted into the container. [Pg.71]

To prepare a given volume of a solution of a given molarity of solute when the weight of the pure, solid solute is to be measured, it is necessary to calculate the grams of solute that are required. This number of moles can be calculated from the desired molarity and the desired volume of solution. [Pg.258]

Grams can be calculated by multiplying moles by the formula weight, as we saw often in Chapters 7 and 8. [Pg.258]


Liquid solute and volume percent Solid solute and weight to volume percent Solid solute and molarity... [Pg.259]


See other pages where Solid Solute and Molarity is mentioned: [Pg.70]    [Pg.258]   


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