All categories

3 years ago

How many grams of Na2SO4 are required to make 0.30 L of 0.500 M Na2SO4?

Chemistry

1 answer:

RSB [31]3 years ago

3 0

Answer: 21.3 g of $Na_2SO_4$ are required to make 0.30 L of 0.500 M $Na_2SO_4$ .

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

moles of $Na_2SO_4$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{142g/mol}$

Now put all the given values in the formula of molality, we get

$0.500M=\frac{xg}{142g/mol\times 0.30L}$

$x=21.3g$

Therefore, 21.3 g of $Na_2SO_4$ are required to make 0.30 L of 0.500 M $Na_2SO_4$ .

You might be interested in

An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle a

Lisa [10]

Answer:

1. 0.0637 moles of nitrogen.

2. The partial pressure of oxygen is 0.21 atm.

Explanation:

1. If we assume ideal behaviour, we can use the Law of ideal gases to find the moles of nitrogen, considering that air composition is mainly nitrogen (78%), oxygen (21%) and argon (1%):

$V_{N_2}=V_{T}\times 0.78=2L \times 0.78 =1.56 L\\PV=nRT\\n_{N_2}=\frac{PV}{RT}=\frac{1 atm\times 1.56 L}{0.0821\frac{atmL}{molK}\times 298 K}\\n_{N_2}= 0.0637 mol$

2. Now, in order to find he partial pressure of oxygen we need to find the total moles of air, and then the moles of oxygen. Then, we use these results to determine the molar fraction of oxygen, to multiply it with total pressure and get the partial pressure of oxygen as follows:

$n_{total}=\frac{1 atm \times 2L}{0.0821 \frac{atmL}{molK}298K}=0.0817 mol$

$V_{O_2}=2L \times 0.21 = 0.42 L\\n_{O_2}=\frac {1atm \times 0.42 L}{0.0821 \frac{atm L}{mol K}298 K}=0.0172 mol\\X_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.0172 mol}{0.0817 mol}= 0.21$

$P_{O_2}=X_{O_2} \times P = 1 atm \times 0.21 = 0.21 atm$

As you see, the molar fraction and volume fraction are the same because of the assumption of ideal behaviour.

3 0

3 years ago

When a solution of 0.1 M Mg(NO3)2 was mixed with a limited amount of aqueous ammonia, a light white, wispy solid was observed, i

Ipatiy [6.2K]

Answer: The net ionic equation is written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of magnesium nitrate and aqueous ammonia (ammonium hydroxide) is given as:

$Mg(NO_3)_2(aq.)+2NH_4OH(aq.)\rightarrow Mg(OH)_2(s)+2NH_4NO_3(aq.)$

A white precipitate of magnesium hydroxide is formed in the above reaction.

Ionic form of the above equation follows:

$Mg^{2+}(aq.)+2NO_3^-(aq.)+2NH_4^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Mg^{2+}(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)$

Hence, the net ionic equation is written above.

8 0

3 years ago

Which of the following statements explains which trial has a lower concentration of the reactant? (5 points) Trial 1, because th

PIT_PIT [208]

Answer:

A)Trial 1 because the average rate of reaction is lower.

Explanation:

I accidentally gave myself low rating my bad

5 0

3 years ago

Read 2 more answers

Identify the type of reaction in KBrO3(s)  KBr(s) + O2(g)

lord [1]

I believe an oxidation reaction is occurring

5 0

3 years ago

Read 2 more answers

how many mL of a 3.5 M sodium hydroxide will be needed to neutralize 15mL of a 4.3 M hydrochloric acid solution?

kobusy [5.1K]

Step 1 : write a valanced equation..
NaOH + HCl 》NaCl + H2O

Step 2 : find the number of mole of HCl..
1000 ml ..contains 4.3 mole
15ml... (4.3÷1000)×15 =...

Stem 3 : use mole ratio....
HCl : NaOH
1 : 1
So mole is same as calculated above...

Step 4 :
3.5 mole of NaOH is in 1000ml
(4.3÷1000)×15 mole is in ....

Do the calculation

6 0

3 years ago