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2 years ago

What is the molality of 30g of NACI dissolved in 350g of water

Chemistry

2 answers:

ivolga24 [154]2 years ago

8 0

Answer:

1.47 m

Explanation:

The molality can be found using the following equation:

Molality = moles solute / mass (kg) solvent

In this case, NaCl is the solute and H₂O is the solvent. To find the molality, you need to (1) convert grams NaCl to moles (via the molar mass), then (2) convert grams H₂O to kilograms, and then (3) calculate the molality.

(Step 1)

Molar Mass (NaCl): 22.990 g/mol + 35.453 g/mol

Molar Mass (NaCl): 58.443 g/mol

30 grams NaCl 1 mole
-------------------------- x ------------------------ = 0.513 moles NaCl
58.443 grams

(Step 2)

1,000 grams = 1 kg

350 grams H₂O 1 kg
-------------------------- x ----------------------- = 0.35 kg H₂O
1,000 grams

(Step 3)

Molality = moles solute / mass (kg) solvent

Molality = 0.513 moles NaCl / 0.35 kg H₂O

Molality = 1.47 m

Anastasy [175]2 years ago

5 0

Answer:

1.45 mol/kg

Explanation:

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How many moles of pcl5 can be produced from 28.0 g of p4 (and excess cl2)?

BARSIC [14]

P₄ + 10Cl₂ ---> 4PCl₅
stoichiometry of P₄ to PCl₅ is 1:4
number of moles of P₄ reacted - 28.0 g / 124 g/mol = 0.22 mol
Cl₂ is in excess therefore P₄ is the limiting reactant, amount of product formed depends on amount of limiting reactant present
according to molar ratio of 1:4
number of PCl₅ moles formed -0.22 mol x 4 = 0.88 mol
0.88 mol of PCl₅ is formed

5 0

3 years ago

A sample of vinegar was found to have an acetic acid concentration of 0.8846 m. What is the acetic acid % by mass? Assume the de

jenyasd209 [6]

Answer:

5.3%

Explanation:

Let the volume be 1 L

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 0.8846*1

= 0.8846 mol

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of CH3COOH,

m = number of mol * molar mass

= 0.8846 mol * 60.05 g/mol

= 53.12 g

volume of solution = 1 L = 1000 mL

density of solution = 1.00 g/mL

Use:

mass of solution = density * volume

= 1.00 g/mL * 1000 mL

= 1000 g

Now use:

mass % of acetic acid = mass of acetic acid * 100 / mass of solution

= 53.12 * 100 / 1000

= 5.312 %

≅ 5.3%

3 0

3 years ago

In the reaction 2H2O (1)+ 2Cl^- (aq)= H2(g)+Cl2 (g)+ 2OH^-(aq), which substance is reduced?

Oksanka [162]

Answer:- C. H

Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.

As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.

The oxidation number in elemental form is zero.

In $H_2O$ , the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in $Cl^-$ is -1. On product side, the oxidation number of hydrogen in $H_2$ is zero and in $OH^-$ the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in $Cl_2$ is 0.