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3 years ago

A solution consists of 44.5 g of silver nitrate, AgNO₃, and 106.5 g water.

Chemistry

1 answer:

katrin2010 [14]3 years ago

5 0

Answer:

Weight percent = 29.5%

Molality = 2.460 mol/kg

Mole fraction = 0.042

Explanation:

The weight percent (%w) of a substance is the mass of it divided by the total mass of the solution, and then multiplied by 100%, so:

%w = [44.5/(44.5 + 106.5)] *100%

%w = 29.5%

The molality (W) of a solution is the number of moles of the solute divided by the mass (in kg) of the solvent. In this case, water is the solvent, and AgNO₃ the solute (water is generally the solvent, and the solvent is always in a larger amount). The number of moles is the mass divided by the molar mass, and the molar mass of AgNO₃ is 169.87 g/mol:

n = 44.5/169.87 = 0.2620 mol

W = 0.2620/0.1065 = 2.460 mol/kg

The mole fraction (x) of a substance is the number of moles of it divided by the total number of moles of the solution. The molar mass of water is 18 g/mol, so the number of moles of water is:

n = 106.5/18 = 5.92 moles

Thus the mole fraction of AgNO₃ is:

x = 0.2620/(0.2620 + 5.92) = 0.042

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Read 2 more answers

Question 1(Multiple Choice Worth 4 points)

zubka84 [21]

Answer

Answer 1 : 28.9 g of CO is needed.

Answer 2 : Six moles of $H_{2}O$ over Nine moles of $O_{2}$

Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Answer 4 : Mass of $O_{2}$ = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Answer 5 : 8.4 moles of sodium cyanide (NaCN) would be needed.

Solution

Solution 1 : Given,

Given mass of $Fe_{2}O_{3}$ = 55 g

Molar mass of $Fe_{2}O_{3}$ = 159.69 g/mole

Molar mass of CO = 28.01 g/mole

Moles of $Fe_{2}O_{3}$ = $\frac{\text{ Given mass of } Fe_{2}O_{3}}{\text{ Molar mass of } Fe_{2}O_{3}}$ = $\frac{55 g}{159.69 g/mole}$ = 0.344 moles

Balanced chemical reaction is,

$Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)$

From the given reaction, we conclude that

1 mole of $Fe_{2}O_{3}$ gives → 3 moles of CO

0.344 moles of $Fe_{2}O_{3}$ gives → 3 × 0.344 moles of CO

= 1.032 moles

Mass of CO = Number of moles of CO × Molar mass of CO

= 1.032 × 28.01

= 28.90 g

Solution 2 : The balanced chemical reaction is,

$2C_{3}H_{6}+9O_{2}\rightarrow 6CO_{2}+6H_{2}O$

From the given reaction, we conclude that the Six moles of $H_{2}O$ over Nine moles of $O_{2}$ is the correct option.

Solution 3 : The balanced chemical reaction is,

$4Fe+3O_{2}\rightarrow 2Fe_{2}O_{3}$

From the given balanced reaction, we conclude that Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Solution 4 : Given,

Given mass of $Zn(ClO_{3})_{2}$ = 150 g

Molar mass of $Zn(ClO_{3})_{2}$ = 232.29 g/mole

Molar mass of $O_{2}$ = 31.998 g/mole

Moles of $Zn(ClO_{3})_{2}$ = $\frac{\text{ Given mass of }Zn(ClO_{3})_{2} }{\text{ Molar mass of } Zn(ClO_{3})_{2}}$ = $(\frac{150\times 1}{232.29})moles$