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3 years ago

How many grams of carbon are in 6.32 moles?

Chemistry

1 answer:

Anit [1.1K]3 years ago

7 0

Answer:

17.18 moles of NaCl are in 2,719 mL of a 6.32

Explanation:

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3 years ago

A solution contains 10.20 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL

AveGali [126]

The question is incomplete, here is the complete question:

A solution contains 10.20 g of unknown compound dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.21°C. The mass percent composition of the compound is 60.98% C , 11.94% H , and the rest is O.

What is the molecular formula of the compound?

Answer: The molecular formula for the given organic compound is $C_6H_{14}O_2$

Explanation:

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

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

Density of water = 1 g/mL

Volume of water = 50.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{50.0mL}\\\\\text{Mass of water}=(1g/mL\times 50.0mL)=50g$

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

$\Delta T_f=i\times K_f\times m$

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 (water) = 0°C

Freezing point of solution = -3.21°C

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

$K_f$ = molal boiling point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute = 10.20 g

$M_{solute}$ = Molar mass of solute = ?

$W_{solvent}$ = Mass of solvent (water) = 50.0 g

Putting values in above equation, we get:

$(0-(-3.21))^oC=1\times 1.86^oC/m\times \frac{10.20\times 1000}{M_{solute}\times 50}\\\\M_{solute}=\frac{1\times 1.86\times 10.20\times 1000}{3.21\times 50}=118.2g$