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neonofarm [45]
3 years ago
14

A sample of glucose ( C6H12O6 ) of mass 8.44 grams is dissolved in 2.11 kg water. What is the freezing point of this solution? T

he freezing point depression constant, Kf , for water is 1.86 °C/mol. (Round your answer to the nearest thousandth)
Chemistry
1 answer:
zhuklara [117]3 years ago
4 0

Answer:

- 0.0413°C ≅ - 0.041°C (nearest thousands).

Explanation:

  • Adding solute to water causes the depression of the freezing point.

  • We have the relation:

<em>ΔTf = Kf.m,</em>

Where,

ΔTf is the change in the freezing point.

Kf is the freezing point depression constant (Kf = 1.86 °C/m).

m is the molality of the solution.

<em>Molality is the no. of moles of solute per kg of the solution.</em>

  • <em>no. of moles of solute (glucose) = mass/molar mass</em> = (8.44 g)/(180.156 g/mol) = <em>0.04685 mol.</em>

<em>∴ molality (m) = no. of moles of solute/kg of solvent</em> = (0.04685 mol)/(2.11 kg) = <em>0.0222 m.</em>

∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.

<em>∴ The freezing point of the solution = the freezing point of water - ΔTf </em>= 0.0°C - 0.0413°C = <em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>

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<span>The principal quantum number (n) is one of four quantum numbers which are assigned to each electron in an atom to describe that electron's state.
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<span>For n = 3, l = 0, 1, 2.</span>

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Option B. A

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Explanation:

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H_2~+~O_2~->~H_2O

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2H_2~+~O_2~->~2H_2O

We have the amount of H_2  and the amount of O_2 . Therefore we have to find the limiting reactive, for this, we have to follow a few steps.

1) Find the moles of each reactive, using the molar mass of each compound (H_2~=~2~g/mol~~O_2=~32~g/mol ).

2) Divide by the coefficient of each compound in the balanced reaction ("2" for H_2 and "1" for O_2).

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