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Law Incorporation [45]

3 years ago

15

Calculate the amount of heat energy required to convert 55.0 g of water at 62.5°C to steam at 124.0°C. (Cwater = 4.18 J/g°C; Cst

eam = 2.02 J/g°C; molar heat of vaporization of liquid water = 4.07 ✕ 104 J/mol)

1 answer:

Grace [21]3 years ago

4 0

Answer:

= 135,647.65 Joules or 135.65 kJ

Explanation:

Heat require to raise the temperature of 55 g water from 62.5°C to 124.0°C will be calculated in steps;

Step 1; heat required to raise water from 62.5°C to 100 °C

Heat = 55 g × 4.18 J/g°C × (100 -62.5)

= 8,621.25 Joules

Step 2; Heat required to convert water to steam without change in temperature;

Heat = moles × Molar Latent heat of vaporization.

= (55g / 18 g/mol)× 4.07 ✕ 104 J/mol

= 12.436 × 10^4 Joules

Step 3: Heat required to raise the temperature of steam from 100°C to 124 °C.

Heat = mass × specific heat capacity × ΔT

= 55 g × 2.02 J/g°C × (124-100)

= 2,666.4 Joules

Therefore; total amount of heat required is;

= 8,621.25 J + 124360 J + 2,666.4 J

<u> = 135,647.65 Joules or 135.65 kJ</u>

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Calculate the freezing point and boiling point of a solution containing 8.15 g of ethylene glycol (C2H6O2) in 96.3 mL of ethanol

pishuonlain [190]

<u>Answer:</u> The freezing point of solution is -117.54°C and the boiling point of solution is 80.48°C

<u>Explanation:</u>

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanol = 0.789 g/mL

Volume of ethanol = 96.3 mL

Putting values in above equation, we get:

$0.789g/mL=\frac{\text{Mass of ethanol}}{96.3mL}\\\\\text{Mass of ethanol}=(0.789g/mL\times 96.3mL)=75.98g$

<u>Calculating the freezing point:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = -114.1 °C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.99°C/m

$m_{solute}$ = Given mass of solute (ethylene glycol) = 8.15 g

$M_{solute}$ = Molar mass of solute (ethylene glycol) = 62 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

$-114.1-\text{Freezing point of solution}=1\times 1.99^oC/m\times \frac{8.15\times 1000}{62g/mol\times 75.98}\\\\\text{Freezing point of solution}=-117.54^oC$

Hence, the freezing point of solution is -117.54°C

<u>Calculating the boiling point:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

Boiling point of pure solution = 78.4°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_b$ = molal boiling point elevation constant = 1.20°C/m.g

$m_{solute}$ = Given mass of solute (ethylene glycol) = 8.15 g

$M_{solute}$ = Molar mass of solute (ethylene glycol) = 62 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-78.4=1\times 1.20^oC/m\times \frac{8.15\times 1000}{62\times 75.98}\\\\\text{Boiling point of solution}=80.48^oC$

Hence, the boiling point of solution is 80.48°C

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Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

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t = time taken by sample = ?

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$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{58.2}{42.8}$

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