All categories

4 years ago

How many grams of a 23.4% by mass naf solution is needed if you want to have 1.33 moles of naf?

1 answer:

6 0

Percentage by mass is the mass of NaF present in 100 g of the solution.
the percentage by mass of NaF is 23.4 %
this means that in 100 g of solution, mass of NaF present is 23.4 g
the number of moles of NaF present - 1.33 mol
mass of NaF - 1.33 mol x 42 g/mol = 55.9 g
when there's 23.4 g of NaF - mass of solution is 100 g
therefore when there's 55.9 g of NaF - mass of solution is 100 / 23.4 x 55.9
= 239 g
mass of solution required is 239 g

You might be interested in

What is the required molarity of a ba(oh)₂ solution to prepare a 1.0 m oh⁻ solution?

timurjin [86]

The molarity of a Ba(OH)2 solution required to prepare a1.0 OH- solution is calculated as follows

write the equation for dissociation of Ba(Oh)2

that is,

Ba(Oh)2 -----> Ba^2+ + 2Oh-

by use of reacting ratio between Ba(Oh)2 to Oh which is 1:2 the molarity of Ba(oh)2 = 1.0/2 = 0.5 M

8 0

4 years ago

Read 2 more answers

To aid in the prevention of tooth decay, it is recommended that drinking water contains 0.900 ppm fluoride (F-). A) How many g o

Romashka-Z-Leto [24]

Answer:

a) <u>1.740 g</u> of F- must be added to a cylindrical water reservoir

b) Grams of sodium fluoride, NaF, that contain this much fluoride:

3.84 g

Explanation:

Step 1. calculate the volume of the tank:

Volume of cylinder =

$\pi r^{2}h$ ,

Here r = radius of the cylinder = d/2

h = depth = 21.80m

$r=\frac{d}{2}$

$=\frac{3.36x10^{2}}{2}$

= 168 m

Volume =

$=\frac{22\times 168^{2}\times 21.80}{7}$

$=1.93\times 10^{6} m^{3}$

2.Convert ppm to g/m3 and Solve for mass of F-

$1ppm = 1g/m^{3}$

$0.9ppm = 0.9g/m^{3}$

Because both ppm and g/m3 are same quantity .

$g/m^{3} =\frac{mass\ of\ F-(g)}{Volume\ m^{3}}\times 10^{6}$

$0.9 =\frac{mass\ of\ F-}{1.93\times 10^{6} m^{3}}\times 10^{6}$

$mass\ of\ F- =1.740g$

mass of F- required = 1.740 g

3. Apply <u>mole concept </u>to calculate grams of sodium fluoride produced

mass of 1 mole of F2 = 38 g

mass of 1 mole of NaF = 42 g

(from periodic table calculate molar mass)

$2Na+F_{2}\rightarrow 2NaF$

Here 1 mole of F2 produce = 2 mole of NaF

So,

38 g of F2 produce = 2 x 42 g of NaF

38 g of F2 produce = 84 g of NaF

1 g of F2 produce = 84/38 g of NaF

1.74 g F2 produce =

$\frac {84}{38}\times 1.74$

1.74 g F2 produce = 3.84 g of NaF

3.84 g of NaF is produced

3 0

3 years ago

Color, texture, and odor are examples of __________ properties. *

zhenek [66]

Answer:

I think the answer is c)

Explanation:

6 0

3 years ago

Read 2 more answers

What are the units of measurement for molarity?

love history [14]

Answer:

Moles per liter.

6 0

3 years ago

Read 2 more answers

Tungsten (W) and chlorine (Cl) form a series of compounds with the following compositions:_______.

deff fn [24]

Answer:

WCl₂, WCl₄, WCl₅, WCl₆

Explanation:

Molar Mass of Tungsten = 184 g/mol

Mass of Chlorine = 35.5 g/mol

In the first compound;

Percentage of tungsten = 72.17 %

Upon solving;

72.17 % = 184

100 % = Total mass

Total mass of compound = 254.95g

Mass of chlorine = 254.95 - 184 = 70.95 (Dividing by 35.35; This is approximately 2 Chlorine atoms.

The Formular is WCl₂

In the second compound;

Percentage of tungsten = 56.45 %

Upon solving;

56.45 % = 184

100 % = Total mass

Total mass of compound = 325.95 g

Mass of chlorine = 325.95 - 184 = 141.95g (Dividing by 35.35; This is approximately 4 Chlorine atoms.

The Formular is WCl₄

In the third compound;

Percentage of tungsten = 50.91 %

Upon solving;

50.91 % = 184

100 % = Total mass

Total mass of compound = 361.42 g

Mass of chlorine = 361.42 - 184 = 177.42 (Dividing by 35.35; This is approximately 5 Chlorine atoms.

The Formular is WCl₅

In the fourth compound;

Percentage of tungsten = 46.39 %

Upon solving;

46.39 % = 184

100 % = Total mass

Total mass of compound = 396.64 g

Mass of chlorine = 396.64 - 184 = 212.64 (Dividing by 35.35; This is approximately 6 Chlorine atoms.

The Formular is WCl₆

4 0

4 years ago