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3 years ago

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a 5.og of sample of compound dissolve in enough water to form 100ml of a solution .this solution form osmotic pressure of 25 tor

r at temperature of 25 .if it is assumed that each molecule of the solution dissociate into two partial what is the molar mass of the solute

1 answer:

storchak [24]3 years ago

7 0

2889 tor plz believe

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Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e

igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

Explanation:

1) Moles of NaCl , $n_1=1 mol$

Mass of water = m= 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute(NaCl)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p=17.19 Torr$

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose , $n_1=1mol$

Mass of water = m = 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p'$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute ( glucose)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p'=17.19 Torr$

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

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Which of the following isotopes is NOT considered to be a major threat to human health from nuclear fallout? a. strontium-90 b.

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Ohhh it’s A) strontium

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3 years ago

Nitrogen gas (112 g) reacts with hydrogen gas to produce 40.8 g of ammonia according to the following

Lemur [1.5K]

Answer:

%yield of NH₃ = 30%

Explanation:

Actual yield of NH₃ = 40.8g

Theoretical yield = ?

Equation of reaction

N₂ + 3H₂ → 2NH₃

Molar mass of NH₃ = 17g/mol

Molarmass of N = 14.00

2 molecules of N = 2 * 14.00 = 28g/mol

Number of moles = mass / molar mass

Mass = number of moles * molar mass

Mass = 1 * 28.00 = 28g of N₂ (the number of moles of N₂ from the equation is 1).

From the equation of reaction,

28g of N₂ produce (2 * 17)g of NH₃

28g of N₂ = 34g of NH₃

112g of N₂ = x g of NH₃

X = (112 * 34) / 28

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Theoretical yield = 136g of NH₃

% yield = (actual yield / theoretical yield) * 100

% yield = (40.8 / 136) * 100

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(5 x 10-21) - (4 x 10-22)

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Answer:

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Explanation:

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Answer

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Explanation:

51/120 = x/360

51(360) = 120x

18360 = 120x

18360/120x = 120x/120x

153 = x

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