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1 year ago

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. The Henry's law constant for helium gas in water at 30 ∘C is 3.7×10−4M/atm; the constant for N2 at 30 ∘C is 6.0×10−4M/atm. a.

If helium gas is present at 2.1 atm pressure, calculate the solubility of this gas. b. If N2 is present at 2.1 atm pressure, calculate the solubility of this gas.2. A solution is made containing 14.7 g of CH3OH in 186 g H2O. a. Calculate the mass percent of CH3OH. b. Calculate the molality of CH3OH.

1 answer:

ludmilkaskok [199]1 year ago

4 0

Explanation:

1) Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$M_{g}=K_H\times p_{g}$

where,

$K_H$ = Henry's constant

$p_{He}$ = partial pressure of gas

a) $K_H=3.7\times 10^{-4} M/atm$

$p_{He}=2.1 atm$

Putting values in above equation, we get:

$M_{He}=3.7\times 10^{-4} M/atm\times 2.1 atm = 7.77\times 10^{-4} M$

The solubility of helium gas is $7.77\times 10^{-4} M$

b) $K_H=6.0\times 10^{-4} M/atm$

$p_{N_2}=2.1 atm$

Putting values in above equation, we get:

$M_{He}=6.0\times 10^{-4} M/atm\times 2.1 atm = 1.26\times 10^{-3} M$

The solubility of nitrogen gas is $1.26\times 10^{-3} M$

2)

a) Mass of solute or methanol , m= 14.7 g

Mass of solvent or water , m'= 186 g

Mass of the solution = M = m + m' = 14.7 g + 186 g = 200.7 g

$w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

$=\frac{14.7 g}{200.7 g}\times 100=7.32\%$

The mass percent of methanol is 7.32%.

b) Molality is defined as moles of solute per kilograms of solvent.

$m=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

Moles of methanol = $\frac{14.7 g}{32 g/mol}=0.4594 mol$

Mass of solvent = 186 g = 0.186 kg

$m=\frac{0.4594 mol}{0.186 kg}=2.4610 mol/kg$

The molality of methanol is 2.4610 mol/kg.

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Nezavi [6.7K]

Answer:

1

The heat of reaction is $Q = 170.92 \ J$

The enthalpy is $\Delta H = -170.91 \ J$

2

The concentration of HCl is $M__{HCl}} =2.468 M$

Explanation:

From the question we are told that

The volume of $NaOH$ is $V_{NaOH} = 45 \ cm^3 = \frac{45} {1000} = 0.045 L$

The number of concentration of $NaOH$ is $M__{NaOH}} = 2M$

The volume of HCl is $V_{HCl} = 20 cm^3$

The number of concentration of $HCl$ is $M__{HCl}} = 1 M$

The temperature difference is $\Delta T = 9.4 ^o C$

Now the heat of reaction is mathematically represented as

$Q = m * c_p * \Delta T$

Where

$c_p$ is the specific heat of water with value $c_p = 4200 J /kg ^o C$

$m = m__{NaOH}} + m__{HCl}}$

Now $m__{NaOH}} = V_{NaOH} * M_{NaOH} * Z_{NaOH}$

where $Z_{NaOH}$ is the molar mass of NaOH with the value of 0.04 kg/mol

So $m__{NaOH}} = 0.045 * 2 * 0.040$

$m__{NaOH}} =0.0036\ kg$

While

$m__{HCl}} = V_{HCl} * M_{HCl} + Z_{HCl}$

Where $Z_{HCl}$ is the molar mass of $HCl$ with the value of 0.03646 kg/mol

$m__{NaOH}} = 0.020 * 1 * 0.03646$

$m__{NaOH}} = 0.000729 kg$

So

$m = 0.0036 + 0.000729$

$m = 0.00433$

=> $Q = 0.00433 * 4200 * 9.4$

$Q = 170.92 \ J$

The enthalpy is mathematically represented as

$\Delta H = - Q$

=> $\Delta H = -170.91 \ J$

From the second question we are told that

The volume of HCl is $V_{HCl_1} = 10cm^3 = \frac{10}{1000} = 0.010 L$

The volume of NaOH is $V_{NaOH_1 } = 30 cm^3 = \frac{30}{1000} = 0.03 L$

The concentration of NaOH is $M_{NaOH} = 2 M$

The first temperature change is $\Delta T = 4.5 ^oC$

The second volume of $V_{HCl_2} = 20 cm^3 = \frac{20}{1000 } = 0.020 m^3$

The mass of NaOH is

$m__{NaOH}} = V_{NaOH} * M_{NaOH} * Z_{NaOH}$

substituting values

$m__{NaOH}} = 0.03 * 2 * 40$

$m__{NaOH}} = 3.6 \ g$

The mass of the product formed is

$m = \frac{Q}{c_p * \Delta T}$

substituting values

$m = \frac{170.91}{4200 * 9} * 1000$

The multiplication by 1000 is to convert it from kg to grams

$m = 4.5 g$

Now the mass of HCl is

$m__{HCl}} = m - m__{NaOh}}$

substituting values

$m__{HCl}} = 4.5 -3.6$

$m__{HCl}} = 0.9 \ g$

Now the concentration of HCl is

$M__{HCl}} = \frac{m_{HCl}}{(Z_{HCl} * *1000) * V_{HCl_1}}}$

The multiplication of $Z_{HCl}$ is to convert it from kg/mol to g/mol

$M__{HCl}} = \frac{0.9}{36.46 * 0.01}}$

$M__{HCl}} =2.468 M$

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1 year ago

Given that the molar mass of NaNO3 is 85.00 g/mol, what mass of NaNO3 is needed to make 4.50 L of a 1.50 M NaNO3 solution?

marissa [1.9K]

Mass = molarity x molar mass( NaNO₃) x volume

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Answer:

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Explanation:

Given data

Added heat (Q): 38.5 J

Change in the temperature (ΔT): 3.0°C

Mass of the metal (m): 17.5 g

Specific heat of the metal (c): ?

We can determine the specific heat of the metal using the following expression.

Q = c × m × ΔT

c = Q / m × ΔT

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Answer :

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Explanation :

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This reaction is a single replacement reaction or displacement in which the the more reactive element (Fe) replace the less reactive element (H).

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