How many grams of pure naoh must be used to prepare 10.0 l of a solution that has a ph of 13? __________g?

Chemistry

1 answer:

padilas [110]3 years ago

8 0

40 g NaOH. You must use 40 g NaOH to prepare 10.0 L of a solution that has a pH of 13.

<em>Step 1</em>. Calculate the pOH of the solution

pOH = 14.00 – pH = 14.00 -13 = 1

<em>Step 2</em>. Calculate the concentration of NaOH

[NaOH] = [OH^(-)] = 10^(-pOH) mol/L = 10^(-1) mol/L = 0.1 mol/L

<em>Step 3</em>. Calculate the moles of NaOH

Moles of NaOH = 10.0 L solution × (0.1 mol NaOH/1 L solution) = 1 mol NaOH

<em>Step 4</em>. Calculate the mass of NaOH

Mass of NaOH = 1 mol NaOH × (40.00 g NaOH/1 mol NaOH) = 40 g NaOH

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Determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. group of answer choices

lilavasa [31]

Considering the ideal gas law and the definition of molar mass, the molar mass of the sample of gas is 9.17 $\frac{g}{mol}$ .

<h3>Ideal gas law</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Molar mass of the sample of gas</h3>

In this case you know:

P= 0.980 arm
V= 1.20 L
T= 287 K
R= 0.082 $\frac{atmL}{molK}$
n= ?

Replacing in the ideal gas law:

0.980 atm× 1.20 L= n× 0.082 $\frac{atmL}{molK}$ × 287 K

Solving:

(0.980 atm× 1.20 L)÷ (0.082 $\frac{atmL}{molK}$ × 287 K)= n

<u><em>0.04997 moles= n</em></u>

On the other hand, you know that the<u><em> mass of the sample of gas</em></u> is <u><em>0.458 grams</em></u>. Replacing in the definition of molar mass:

$molar mass=\frac{0.458 grams}{0.04997 moles}$