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3 years ago

which of these solutions has the lowest freezing point? 0.25 m nacl 0.5 m nacl 1.0 m nacl 1.5 m nacl 2.0 m nacl

Chemistry

2 answers:

zloy xaker [14]3 years ago

6 0

Answer:

2.0 m nacl

Explanation:

Depression of freezing point is given by the equation:

ΔTf = m. Kf. i

where i is the Van't Hoff factor, m is the molality, and Kf is the freezing point depression constant.

As per the formula, depression of freezing point is directly proportional to the molality. 2.0 m NaCl has the highest molaity and will have the maximum depression in freezing point. As a result it will also have the lowest freezing point.

Alinara [238K]3 years ago

4 0

The more particles (ions or molecules) that you can put into solution, the lower the freezing point.

the answer is E. 2.0 M nacl

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Answer:

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Explanation:

Here Pressure in a container is given as

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Here

P is the pressure which is to be calculated

ρ is the density of the gas which is to be calculated as below

$\rho =\frac{mass}{Volume of container}$

Here

mass is to be calculated for 75 gas phase molecules as

$m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times 10^{-21} g$

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So density is given as

$\rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3$

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$=RMS^2$

Here RMS is the Root Mean Square speed given as 605 m/s so

$=RMS^2\\=(605)^2\\=366025$

Substituting the values in the equation and solving

$P=\frac{1}{3} \rho \\P=\frac{1}{3} \times 7.97 \times 10^{-21} \times 366025\\P=9.72 \times 10^{-16} Pa$

So the average pressure in the container due to these 75 gas molecules is $P=9.72 \times 10^{-16} Pa$

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