All categories

3 years ago

What mass of Sodium Chloride is required to make 100.0 mL of 3.0 M solution?

Chemistry

1 answer:

Julli [10]3 years ago

5 0

Answer:

17.55 g of NaCl

Explanation:

The following data were obtained from the question:

Molarity = 3 M

Volume = 100.0 mL

Mass of NaCl =..?

Next, we shall convert 100.0 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

100 mL = 100/1000

100 mL = 0.1 L

Therefore, 100 mL is equivalent to 0.1 L.

Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:

Molarity = 3 M

Volume = 0.1 L

Mole of NaCl =?

Molarity = mole /Volume

3 = mole of NaCl /0.1

Cross multiply

Mole of NaCl = 3 × 0.1

Mole of NaCl = 0.3 mole

Finally, we determine the mass of NaCl required to prepare the solution as follow:

Mole of NaCl = 0.3 mole

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =?

Mole = mass /Molar mass

0.3 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 0.3 × 58.5

Mass of NaCl = 17.55 g

Therefore, 17.55 g of NaCl is needed to prepare the solution.

You might be interested in

How many moles of brcl form when 2.74 mol cl2 react with excess br2?

egoroff_w [7]

Following reaction arise between Br2 and Cl2

Br2 + Cl2 → 2BrCl
(1mole) (1mole) (2moles)

From above balanced reaction, it can be seen that 1 mole of Br2 reacts with 1 mole of Cl2 to form 2 mole of BrCl

Thus, when <span>2.74 mol Cl2 reacts with excess Br2, 2.74 X 2 = 5.48 moles of BrCl will be formed. </span>

5 0

3 years ago

The value of Kp for the reaction NO(g) 1 1 2 O2(g) 4 NO2(g) is 1.5 3 106 at 25°C. At equilibrium, what is the ratio of PNO2 to P

expeople1 [14]

Answer : The ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$

Solution : Given,

$K_p=1.5\times 10^6$

$p_{O_2}$ = 0.21 atm

The given equilibrium reaction is,

$NO(g)+\frac{1}{2}O_2\rightleftharpoons NO_2(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{NO_2})}{(p_{NO})\times (p_{O_2})^{\frac{1}{2}}}$

Now put all the given values in this expression, we get:

$1.5\times 10^6=\frac{(p_{NO_2})}{(p_{NO})\times (0.21)^{\frac{1}{2}}}$

$\frac{(p_{NO_2})}{(p_{NO})}=(1.5\times 10^6)\times (0.21)^{\frac{1}{2}}$

$\frac{(p_{NO_2})}{(p_{NO})}=6.87\times 10^5$

Therefore, the ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$

5 0

3 years ago

How many moles are in 4.5 g of<br> Sodium Chloride, NaCl?

Sergeu [11.5K]

Answer:

The answer to the question is 0.07 moles

6 0

2 years ago

What is the nuclear equation for decay of Am - 241

adelina 88 [10]

Americium is a radioactive element .

It goes through alpha decay during nuclear reactions .

The nuclear reaction.is given by

$\rm {}^{241}_{95}Am\longrightarrow {}^{237}_{93}Np+{}^4_2\alpha$

Np stands for Nuptunium

Here alpha can be written as He too

3 0

1 year ago

Write the formula for potassium oxide. why do you not need prefixes in the name

Molodets [167]

Potassium oxide: K₂O.

There's no need for prefixes since K₂O is an ionic compound.

<h3>Explanation</h3>

Find the two elements on a periodic table:

Potassium- K- on the left end of period four.
Oxygen- O- near the right end of periodic two.

Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.

Potassium is a metal,
Oxygen is a nonmetal.

A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:

Potassium ions: Potassium is group 1 of the periodic table. It is an alkaline metal. Like other alkaline metals such as sodium Na, potassium K tends to lose one electron and form ions of charge +1 in compounds. The ion would be K⁺.

Oxide ions from oxygen: Oxygen is the second most electronegative element on the periodic table. It tends to gain two electrons and form the oxide ion $\text{O}^{2-}$ when it combines with metals.

The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each $\text{O}^{2-}$ ion is twice that on a $\text{K}^{+}$ ion. Each $\text{K}^{+}$ would pair up with two $\text{O}^{2-}$ . Hence the subscript in the formula: $\text{K}_{\bf 2}\text{O}$ .

There are two classes of compounds:

Covalent compounds, which need prefixes, and
Ionic compounds, which need no prefix.

Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide $\text{CO}_2$ , the prefix di- indicates that there are two oxygen atoms in the formula $\text{CO}_2$ . However, there's no need for prefix in ionic compounds such as $\text{K}_2\text{O}$ .

7 0

3 years ago